How Do Time Dilation and Space Expansion Affect Gravitational Redshift?

[mysearch]

Based on the quote below and the link to Wikipedia, I believe the implication is that gravitational redshift will occur when a photon climbs out of the gravity well, while blueshift will occur when falling into the gravity well. However, I am trying to clarify whether both time dilation and space expansion have to be taken into consideration when an observer measures the frequency or wavelength of a photon at different positions in the gravity well.

"[URL

Gravitational Redshift:
When a photon is emitted by a given process, for example a transition of an electron from one specific energy level to another specific level in a specific element the photon is emitted at an intrinsic frequency irrespective of where the element is located in the gravity well. Say the photon is moving upwards, then measurements made further up, by clocks with faster coordinate time make it appear as if the frequency of the photon is getting slower but the coordinate frequency (and energy) remains constant and it is the wavelength (and coordinate speed of light) that is increasing. You could view the increasing wavelength as a gravitational length contraction effect. Local observers progressively further up the gravity well see the frequency as slowing down and so to them the photon appears to be losing energy and the wavelength increasing which is consistent with the local speed of light being constant for any observer. Depends on your point of view.

Not sure as to what is `intrinsic` to frequency as I would have thought frequency is subjective to an observer?

If the observer is collocated with the photon source within a gravity well, does local time define the frequency and therefore the energy E=hf?

To a distant observer, far from the gravity well, time ticks faster, so would the photon, when arriving, appear to have a lower frequency and energy?

However, along the way, space would have contracted as space is said to expand in the direction of increasing gravity. Therefore, would this effect cause the wavelength to shorten?

If c=f\lambda and [c] is constant, would the shorten wavelength not cause frequency to increase and cancel the effect of time dilation?​

Clearly theory suggests that this is not the case, but then implies there is a different perception of the energy associated with the photon. Therefore, would welcome any clarification of these issues.

 

[Mentz114]

However, along the way, space would have contracted as space is said to expand in the direction of increasing gravity. Therefore, would this effect cause the wavelength to shorten?

I think it would be the other way round. The wavelength would appear longer.

For light, frequency and wavelength are not independent, and so you can regard a red-shift either as an increase in wave length or a reduction in frequency.

Red-shifted light will deliver less energy. This is not a surprise because energy is not relativistically invariant.

Have you come across the Pound-Rebka experiment ? I guess you have since you quote Prof. Wiki.
en.wikipedia.org/wiki/Pound-Rebka_experiment

 

[pmb_phy]

Based on the quote below and the link to Wikipedia, I believe the implication is that gravitational redshift will occur when a photon climbs out of the gravity well, while blueshift will occur when falling into the gravity well. However, I am trying to clarify whether both time dilation and space expansion have to be taken into consideration when an observer measures the frequency or wavelength of a photon at different positions in the gravity well.

All that is neccesary is that time dilation is present, i.e, that g₀₀ is a function of position. g₀₀ depending on position does not imply that space is altered.

Pete

 

[mysearch]

Mentz114: Thanks for the comments. I have attached a response against each below:

I think it would be the other way round. The wavelength would appear longer.

As you approach a large mass [M] doesn’t time tick more slowly and the radial distance [r] expand? If so, I was assuming a photon traveling away from [M] would perceive a contraction in unit space and therefore the wavelength would shorten, although this doesn’t seem to be the case.

For light, frequency and wavelength are not independent, and so you can regard a red-shift either as an increase in wave length or a reduction in frequency.

I agree based on the assumption that c=f\lambda and [c] is constant. This was why I was asking about the combined effects of time dilation and space contraction when the photon travels away from [M]. Pete in post #3 appears to be implying that time dilation alone causes the gravitational shift, which suggests that the frequency is fixed at N cycles per second, when initially emitted, but as the photon moves out of the gravity well, N cycles per second appears lower because time is ticking faster. What I didn’t understand was why the radial contraction per unit length didn’t not shorten the wavelength, i.e. increase frequency, and therefore cancel the effects of time dilation. Clearly, experiments suggest this is not the case, but I couldn’t really explain to myself why both time and space effects didn’t apply.

Red-shifted light will deliver less energy. This is not a surprise because energy is not relativistically invariant.

How does the conservation of energy get resolved if the photon appears to lose energy? In classical physics a physical mass can lose kinetic energy, but its total energy is conserved in its potential energy.

Have you come across the Pound-Rebka experiment ? I guess you have since you quote Prof. Wiki.

Actually I hadn’t, so thanks for the reference.

Pmb_phy/Pete: Thanks for the feedback, unfortunately I not that familiar with the meaning of g_{00} notation. As I have tried to explain above, I don’t really understand why the changes in both time and space don’t cancel out. Most texts seem to simply quote the net effect without reference to the underlying physical cause, e.g.

In physics, light of a certain wavelength originating from a source placed in a region of stronger gravitational field will be found to be of longer wavelength when received by an observer in a region of weaker gravitational field.

You seem to suggest that time dilation causes the photon to have a lower frequency, i.e. longer wavelength, and that the effects of radial space contracting do not cause the wavelength to shorten, i.e. higher frequency.

 

[Mentz114]

What I didn’t understand was why the radial contraction per unit length didn’t not shorten the wavelength

The speed of light would have to change to accommodate a lower frequency with a shorter wavelength.

Pete is right, only g(r)_{00} is used to calculate gravitational redshift. The radial component is irrelevant. This must be in the nature of light. A mass thrown up can lose velocity, light cannot.

 

[mysearch]

Request for confirmation

Pete is right, only g_{00} is used to calculate gravitational redshift. The radial component is irrelevant. This must be in the nature of light. A mass thrown up can lose velocity, light cannot.

As indicated earlier I have not really studied the vector-tensor maths associated with g_{00}, however looking around on the web I found one reference that suggested that g_{00} is associated with time dilation where:

d\tau = \sqrt{g_{00}} dt

This suggests that in the context of gravity only, i.e. no relative velocity, that g_{00} might be related to the relativity factor [\gamma], e.g.

g_{00} = 1-Rs/r??

I mention this simply because I am interested in trying to resolved some physical interpretation that explains the effects under discussion. Therefore, I wanted to try to expand on your other comment:

The speed of light would have to change to accommodate a lower frequency with a shorter wavelength.

While possibly over simplistic, I have described relativity in terms of its separate effects on space and time. Special relativity describes the effect of velocity, which remains invariant because time dilation and space contraction are equal and maintain constant velocity in both frames of reference. General relativity incorporates the additional effect of gravity, which also leads to time dilation, although space expands in the radial direction. As such, velocity cannot be invariant. As such, velocity is simply a concept of distance traveled in a given time, i.e. v=d/t, and this is all c=f\lambda would appear to be saying. However, the assumption that v=d/t may be the issue that has misled me, because in the context of wave propagation, the frequency [f] is defined by the energy [E=hf] and the propagation velocity [v=c] is defined by the media, i.e. vacuum permittivity & permeability. As such, it might be more accurate to say that wavelength results from \lambda=c/f.

I not sure whether this is an appropriate physical interpretation, but it suggests that the wavelength of a photon, as measured by a distant observer, can only result from changes in [c] and [f]. If [c] is a constant in each frame of reference, then only the time dilation effects on [f] lead to the gravitational redshift observed by experiments. As such, the radial contraction that occurs when moving out of the gravity well would not directly affect wavelength as I had originally assumed. Is this a valid conclusion?

 

[MeJennifer]

I not sure whether this is an appropriate physical interpretation, but it suggests that the wavelength of a photon, as measured by a distant observer, can only result from changes in [c] and [f]. If [c] is a constant in each frame of reference, then only the time dilation effects on [f] lead to the gravitational redshift observed by experiments. As such, the radial contraction that occurs when moving out of the gravity well would not directly affect wavelength as I had originally assumed. Is this a valid conclusion?

The frequency the distant observer measures depends on the curvature differential between the event of emission and absorption. In other words, f does not change in a gravitational field but the measuring apparatus changes in such a field. A measured blue shifted photon is really a photon observed by a red shifted apparatus.

 

[mysearch]

Response to post #7

I accept that we could describe the gravitational redshift in terms of a differential curvature of spacetime. However, I would have thought this curvature would still correspond to the space and time effects relative to 2 positions in the gravity well. As far as I was aware, time dilation is a real effect against which we then subsequently define frequency. In this context, the real change in the perception of time appears to be more fundamental than the knock-on effects on the measuring apparatus. This is not being forwarded as a statement fact, simply the assumptions to which I was working

 

[Mentz114]

mysearch,

going back to an earlier remark,

As I have tried to explain above, I don’t really understand why the changes in both time and space don’t cancel out.

Well, why should they ?

Most texts seem to simply quote the net effect without reference to the underlying physical cause

The 'cause' is the difference in gravitational field strength. This is measured by g_00. Actually it's a bit more subtle but without a full understanding of curved space, hard to explain.

This agrees with experiment. If you're looking for an explanation in terms of underlying structures, then you're into uncharted territory.

In post #6 you're making a mistake by trying to make analogies between the gamma of SR and gravitational effects. SR is formulated in the Minkowski space-time, GR extends this to a general spacetime where curved worldlines can be caused by geometry alone ( gravity) whereas in SR, a curved worldline always implies an external force. The conclusions of SR do not carry over to GR in a simple way.

One more thing. Suppose we have observers A and B, and A sends some light to B who measures the frequency. It is not possible to say whether any difference observed is a result of an 'intrinsic' change in the light or a difference in the measuring apparatus, caused by transporting it to another place. No experiment can distinguish these situations.

 

[mysearch]

Response to #9

The 'cause' is the difference in gravitational field strength. This is measured by g_00. Actually it's a bit more subtle but without a full understanding of curved space, hard to explain. This agrees with experiment. If you're looking for an explanation in terms of underlying structures, then you're into uncharted territory.

A fully accept that I am on a learning curve and for those who have studied this subject in detail, it must be difficult to simplify some concepts to basic principles. It is true that I was hoping to rationalise, at least to myself, the basic effects of relativity in terms of time dilation and space expansion/contraction.

In post #6 you're making a mistake by trying to make analogies between the gamma of SR and gravitational effects.

This wasn’t my intention; I was simple reproducing an inference in
https://www.physicsforums.com/archive/index.php/t-125057.html

It is not possible to say whether any difference observed is a result of an 'intrinsic' change in the light or a difference in the measuring apparatus, caused by transporting it to another place. No experiment can distinguish these situations.

This is a very good point and one also argued in post #7. However, I thought it was an axiom of relativity that everything is relative. As such, it seems to be a philosophical point as to whether the photon changes in transit or the observer/measuring apparatus changes with respect to the photon - see rationale below. From the perspective of the receiving observer, the photon has less energy than a photon from a comparable source emitted locally.

I accept that my attempt to rationalise an explanation in #6 may appear naïve, but it appeared to help clarify some important aspects of gravitational redshift. First, \lambda=c/f seem to help explain why only time dilation affected frequency. Equally, it appears to provide a practical interpretation of the energy lost associated with a photon moving away from a gravitational mass. Based on the conservation of energy, this lost energy has to be accounted in some other form. In classical physics, the energy associated with a mass [m] can be defined in terms of its kinetic (Ek) and potential (Ep) energy that in-turn is linked to velocity and gravitation:

Total Energy (Et) = Ek + Ep =1/2mv^2 + (-GMm/r)

At face value, this is more difficult for a photon as it is said to have no rest mass [m0]. However, if we link Planck’s equation (E=hf) with Einstein’s equation (E=mc^2), it suggests that photons have kinetic mass (mk= hf/c^2). Now if time dilation causes frequency [f] to reduce, in the case of an outgoing photon, it also causes the photon to lose energy. However, the photon now has more potential to gain energy, i.e. increase in frequency, by falling back to a lower position in the gravity well. As such, mass and frequency act as equivalent concepts against which the total energy can be accounted in any frame of reference. It just seems easier to me to assign the changes to the photon coming into the observers frame of reference rather than assume the photon is unchanged, which then requires entire frame of reference to be `recalibrated` to some other arbitrary, but remote frame of reference.

The only problem I have with the localise perspective is that I seems to suggest that the energy of a photon falling into the event horizon would be infinitely blueshifted, implying infinite energy. Again, I am not making statement of fact here, simply outlining some assumptions on which I am attempting to learning more about this subject, so have much appreciated the help and feedback.

 

[yuiop]

I accept that we could describe the gravitational redshift in terms of a differential curvature of spacetime. However, I would have thought this curvature would still correspond to the space and time effects relative to 2 positions in the gravity well. As far as I was aware, time dilation is a real effect against which we then subsequently define frequency. In this context, the real change in the perception of time appears to be more fundamental than the knock-on effects on the measuring apparatus. This is not being forwarded as a statement fact, simply the assumptions to which I was working

Consider the following example. There are 3 observers (A,B,C) at various heights in a gravity well. According to an observer at infinity the gravitational gamma factor g= 1/ \sqrt{1-Rs/R} is 8, 4 and 2 for A,B and C respectively with A being the deepest in the well. Say a photon is emitted with a frequency of 1 and wavelength of 1 at location A as measured by A. The measurements of (frequency, wavelength, speed of light) made by the observer (D) at infinity at locations A, B, C, and D would be:

A' = (1/8, 1/8, 1/64)
B' = (1/8, 1/2, 1/32)
C' = (1/8, 2, 1/4)
D' = (1/8, 8, 1)

and the local measurements would be:

A = (1, 1, 1)
B = (1/2, 2, 1)
C = (1/4, 4, 1)
D = (1/8, 8, 1)

In other words the coordinate measurements according to the observer at infinity indicate the frequency remains constant and wavelength get longer by a factor of gamma squared and the coordinate speed of light also increases by a factor of gamma squared as the photon climbs out the well. To the observer at infinity the energy of the photon is constant.

To the local observers each observer measures a lower frequency and a longer wavelength locally than the observers further down and the same value for the local speed of light c.

That is my analysis that is consistent with the Schwarzschild metric. As with special relativity, it is not possible to state which observers point of view is the more "fundemental".

 

[yuiop]

...

As you approach a large mass [M] doesn’t time tick more slowly and the radial distance [r] expand?

The radial distance effect is the other way round. Approaching a large mass, clocks ticks more slowly by gamma and rulers contract by gamma and the coordinate speed of light slows down by gamma squared, but the combined effects result in the local speed of light being constant for all observers.

 

[mysearch]

Kev, thanks for the comments and especially the analysis in #11, as it is always helpful to look at the issues in terms of a practical example. I was in the process of working through your example, but then saw post #12, which has thrown up some doubts in my mind.

I accept I am no expert, but I was pretty sure that the radial distance expands under gravity as [r] approaches [Rs], i.e. its the opposite to special relativity, in which length contracts as [v->c]. This is why the (1-Rs/r) is the denominator in the [dr] term in the Schwarzschild metric.

It will probably be tomorrow before I can respond properly. Thanks.

 

[yuiop]

Kev, thanks for the comments and especially the analysis in #11, as it is always helpful to look at the issues in terms of a practical example. I was in the process of working through your example, but then saw post #12, which has thrown up some doubts in my mind.

I accept I am no expert, but I was pretty sure that the radial distance expands under gravity as [r] approaches [Rs], i.e. its the opposite to special relativity, in which length contracts as [v->c]. This is why the (1-Rs/r) is the denominator in the [dr] term in the Schwarzschild metric.

It will probably be tomorrow before I can respond properly. Thanks.

Which of these statements do you disagree with?

a) Gravitational time dilation slows clocks by the gamma factor.
b) Coordinate speed of light slows by gamma squared.
c) Local speed of light is always c.

If you agree with statements a,b and c you have to come to the conclusion that length (rulers) have to contract by gamma aproaching a massive body.

The confusion may be the difference between "radial distance" for which there is no simple conversion factor and the length of local almost infinitesimal ruler.

 

[mysearch]

Response to Kev #11, 12 & 14

Hi Kev
Saw your last post, so will start with the last point you raised:

The confusion may be the difference between "radial distance" for which there is no simple conversion factor and the length of local almost infinitesimal ruler.

This could well be the case, but I will try to explain my `frame of reference` then you can tell me if we are talking at cross-purposes. I have no problem with bullets (a) or (c), so maybe the confusion is with respect to bullet (b). I am only providing the following simplified form of the Schwarzschild metric as a reference:

c^2 d\tau^2=c^2\left(1-Rs/r\right)dt^2 - \frac{dr^2}{\left(1-Rs/r\right) }

Without resorting to any derivation, the placement of the (1-Rs/r) terms suggests that the relativistic effects, under gravity, works in different direction for time and space. Also the equations for proper time [d\tau] and proper distance [ds], given in various sources, appears to support this suggestion:

d\tau = \sqrt{\left(1-Rs/r\right)}dt

ds = \frac {dr}{\sqrt{\left(1-Rs/r\right)}}

According to the Schwarzschild metric, the proper radial distance, the actual distance measured by an observer at rest at radius r, between two spheres separated by an interval dr of circumferential radius r is \left(1-Rs/r\right)^{-1/2}dr, which is larger than the radial interval dr expected in a flat, Euclidean geometry
http://casa.colorado.edu/~ajsh/schwp.html

Before continuing, let me just clarify a few assumptions. I am assuming that the local observer/apparatus measures 1 second per second and 1 metre per metre. Therefore, the time and space effects are all relative to a distant observer, e.g. (D) observing (A). I am also assuming that both the local and distant observers are stationary, so no velocity effects. What I not too sure about is that if the local observer (A), in curved space around a black hole, measures the circumference, unaffected by gravity, and then divides by 2\pi, would the radius calculated and the actual radius observed differ and allow the local observer to recognise he was in curved space?

Anyway, returning to your example, albeit in a slight different context, but using your values. The distant observer (D) is `conceptually` watching a `hypothetical` experiment in (A). The experiment consists of a photon being fired between 2 points, aligned on the radial axis, separated by 1 light-second. As the photon is fired, the source flashes a light, as does the receiver when the photon is received. In the context of (A), the flashes are separated by 1 second, as the local speed of light is always [c].

However, following your example, time in (D) is ticking 8x faster than (A) and therefore we might conclude the observed frequency was 1/8 as it is inversely proportional to time. Again, this aligns to your example. However, this would imply that the time, as measured by (D), between the flashes was 8 seconds. Of course, if we go with the previous assumption about radial space expanding on approaching [Rs], then observer [D] would measure the distance between source and receiver as 8 light-seconds, which would imply the photon was still traveling at [c]. I fully accept this logic may be flawed, but would like to understand why.

Just a few other points I want to highlight. When I originally raised the question about gravitational redshift, based on the assumptions above, it was because I couldn’t resolve why the gravitational effects on time and space didn’t cancel out. If time slows approaching [Rs], then it was logical to assume that the emitted frequency of a photon would be lower, in-line with experimental observation. However, if radial space contracts as the photon travels from (A) out to (D), then I initially assumed this would affect the wavelength, which in-turn would affect the frequency based on c=f\lambda. However, I thought this might be resolved based on the fact that the physics of wave propagation only states the frequency [f] is defined by the energy [E=hf], whereas the propagation velocity [v=c] is defined by the media of propagation, i.e. vacuum permittivity and permability. As such, the wavelength \lambda=c/f is determined by [c] and [f]. If [c] is constant in all local frames of reference, then the gravitational redshift would correspond to experiment and align to the effects of time dilation only. Again, I fully accept this might all be unsubstantiated speculation on my part, but my goal was to try to provide some physical interpretation of the relativistic effects. Many Thanks

 

[yuiop]

Hi mysarch,

there is quite a lot in your post so I will address the first few points only for now and come back to the rest as time allows.

... I have no problem with bullets (a) or (c), so maybe the confusion is with respect to bullet (b). I am only providing the following simplified form of the Schwarzschild metric as a reference:

c^2 d\tau^2=c^2\left(1-Rs/r\right)dt^2 - \frac{dr^2}{\left(1-Rs/r\right) }

For light dtau is zero. So we can say for a falling photon:


0 = c^2\left(1-Rs/r\right)dt^2 - \frac{dr^2}{\left(1-Rs/r\right) }

Rearrange and simplify:

\frac{dr^2}{\left(1-Rs/r\right) } = c^2\left(1-Rs/r\right)dt^2

\frac{dr^2}{dt^2} = c^2 \left(1-Rs/r\right)^2

\frac{dr}{dt} = c \left(1-Rs/r\right)

.. proving my assertion in bullet point b that the vertical coordinate speed of a falling photon in a gravitational field (as measured by an observer at infinity) is inversely proportional to the gravitational gamma factor squared.

Without resorting to any derivation, the placement of the (1-Rs/r) terms suggests that the relativistic effects, under gravity, works in different direction for time and space. Also the equations for proper time [d\tau] and proper distance [ds], given in various sources, appears to support this suggestion:

d\tau = \sqrt{\left(1-Rs/r\right)}dt

ds = \frac {dr}{\sqrt{\left(1-Rs/r\right)}}

the transformation for time and space in SR are in "different directions" too. Proper time is less than the time measured in the primed frame and proper length is greater than the length measured in the primed frame.

dt' = dt \sqrt{\left(1-v^2/c^2)}

dx' = \frac {dx}{\sqrt{\left(1-v^2/c^2)}}

 

[mysearch]

Response to Kev #16

Realise you will need to consider my comments in #15. However, wanted to raise some comments against your initial response. I agree with your basic derivation of the observed photon velocity from (D), although I added the \pm to reflect the quadratic solution:

Photon: \frac{dr}{dt} = \pm c \left(1-Rs/r\right)

Simply for reference, here are some other solutions, starting with the observed free-fall velocity from (D):

Free-Fall Mass: \frac{dr}{dt} = c \left(1-Rs/r\right) \sqrt{Rs/r}

The implication of this equation is that both velocities perceived by (D) collapse to zero at [Rs], presumably corresponding to time slowing to a halt at [Rs]. Of course, solutions with respect to proper time [d\tau] suggest otherwise:

Photon: \frac{dr}{d\tau} = -c \sqrt{Rs/r} \pm c

Free-Fall Mass: \frac{dr}{d\tau} = -c \sqrt{Rs/r}

Now the speed of light [c] appears to maintain its value [c] with respect to the local observer (A), even when (A) is free-falling at [c]. To be honest I have never been able to resolve both outcomes from the Schwarzschild metric. At one level, time dilation with respect to (D) appears to be supported by experiment, but leads to a frozen star rather than a black hole. While the other solution, even taking into account arguments concerning coordinate singularity, seems to proceed through the event horizon but cannot seem to relate time back to (D). However, as far as I know, a series of local measurements in (A), (B) and (C) would all measure [c] as [c] and suggest the photon frequency was affected by the initial time dilation.

Sorry, I didn’t understand where your last 2 equations come from, as they do not seem to correspond to the Lorentz transforms:

t' = \gamma \left( t - \frac{v x}{c^{2}} \right)\rightarrow \gamma (t) when x=0

x' = \gamma \left( x - v t \right) \rightarrow \gamma (x) when t=0

Apologises for raising so many issues, but I would be interested how your position explains gravitational redshift. Again, many thanks.

 

[yuiop]

...

The implication of this equation is that both velocities perceived by (D) collapse to zero at [Rs], presumably corresponding to time slowing to a halt at [Rs]. Of course, solutions with respect to proper time [d\tau] suggest otherwise:

Photon: \frac{dr}{d\tau} = -c \sqrt{Rs/r} \pm c

Free-Fall Mass: \frac{dr}{d\tau} = -c \sqrt{Rs/r}

Now the speed of light [c] appears to maintain its value [c] with respect to the local observer (A), even when (A) is free-falling at [c]. To be honest I have never been able to resolve both outcomes from the Schwarzschild metric. At one level, time dilation with respect to (D) appears to be supported by experiment, but leads to a frozen star rather than a black hole. While the other solution, even taking into account arguments concerning coordinate singularity, seems to proceed through the event horizon but cannot seem to relate time back to (D).

One difficulty with using the Schwarzschild solution at exactly r=R_s is that no physical observer could remain stationary at the event horizon so the equztion is not physical at that exact point. However, there are other issues. In another thread we showed that the interior Schwarzschild solution shows that time reverses at certain coordinates below the Schwarzschild radius even before a black hole is fully formed (collapsed to a singularity) suggesting that physical process (time related) prevents a singularity forming and all the mass accumulates in the thin shell just outside the event horizon. This is consistent with the frozen star viewpoint. It is also consistent with the Hawking entropy formula that shows the entropy is proportional to the surface area of the black hole event horizon. Charge and mass accumulated on such a shell would appear to anyone outside the shell to be from a point source just as the gravitational effect of the Earth above the surface is mathematically the same as having all the mass located at the centre even though we know it is not really all located there.

However, as far as I know, a series of local measurements in (A), (B) and (C) would all measure [c] as [c] and suggest the photon frequency was affected by the initial time dilation.

That is exactly what I stated the local observers at A, B and C would measure. Is there measurement the only valid viewpoint? I think not. For example the local observers in Special Relativity can not measure length contraction and time dilation of there own rulers and clocks but that does not mean length contraction and time dilation do not exist. Similarly, the two observers with relative velocity each measure the others clock to ticking slower than their own clock and we know that each has a valid viewpoint even thought they seem contradictory.

Sorry, I didn’t understand where your last 2 equations come from, as they do not seem to correspond to the Lorentz transforms:

t' = \gamma \left( t - \frac{v x}{c^{2}} \right)\rightarrow \gamma (t) when x=0

x' = \gamma \left( x - v t \right) \rightarrow \gamma (x) when t=0

x is the coordinate of a point. Length would be measured as the difference of two coordinates such as x2-x1.

For length:

(x_2' - x_1')/\gamma = \left( x_2 - v t \right) \right - \left( x_1 - v t \right)

(x_2' - x_1')/\gamma = \left( x_2 - x_1 \right)

(x_2' - x_1') = \left( x_2 - x_1 \right)\gamma

For time:

(t_2' -t_1') = \gamma \left( t_2 - \frac{v x_2}{c^{2}}\right) -\gamma \left( t_1 - \frac{v x_1}{c^{2}}\right)

Since x2 and x1 are not the same for non zero t we use the reverse transformation:

t = \gamma \left( t' + \frac{v x}{c^{2}} \right)

(t_2 -t_1) = \gamma \left( t_2' + \frac{v x_2'}{c^{2}}\right) -\gamma \left( t_1' + \frac{v x_1'}{c^{2}}\right)

Now we can say x2' = x1' for the clock in the moving frame and it follows:

(t_2 -t_1) = \gamma \left( t_2' - t_1' \right)

(t_2' -t_1') = \left( t_2 - t_1 \right)/\gamma

Apologises for raising so many issues, but I would be interested how your position explains gravitational redshift. Again, many thanks.

Well, my position is that there is no single explanation. From the point of view of local observers frequency of the photon changes due to time dilation and the speed of light is constant. From the point of view of an observer at infinity (or at any constant non local location) the frequency remains constant, the wavelength is stretched as the photon rises and the speed of light increases. Both points of view are equally valid.

 

[sirzerp]

I have a different question related to Gravitational Redshift.

We know that the universe is loosing mass as the stars produce photons. We also know that photons are red shifted when moving away from a star, and blue shifted when moving to a star.

Let’s do a little thought experiment. If a photon leaves the sun at year zero and is red shifted as it leaves, and then travels a 100 million years and is reflected back; and then travels another 100 million years. When the same photon that left 200 million years returns to the sun, at what wavelength is it at?

That sounds contrived, but I’m trying to ask, how is mass lost by a star, related to red shifting of same photon created by the star, when reflected and lack of blue shifting when returning to the star.

As the solar mass decreases, does that mean all reflected photons back to that star system have to return red shifted?

Anyone have the wavelength math for a one percent mass decrease and 200 million year round trip?

 

[yuiop]

I have a different question related to Gravitational Redshift.

We know that the universe is loosing mass as the stars produce photons. We also know that photons are red shifted when moving away from a star, and blue shifted when moving to a star.

Let’s do a little thought experiment. If a photon leaves the sun at year zero and is red shifted as it leaves, and then travels a 100 million years and is reflected back; and then travels another 100 million years. When the same photon that left 200 million years returns to the sun, at what wavelength is it at?

That sounds contrived, but I’m trying to ask, how is mass lost by a star, related to red shifting of same photon created by the star, when reflected and lack of blue shifting when returning to the star.

As the solar mass decreases, does that mean all reflected photons back to that star system have to return red shifted?

Anyone have the wavelength math for a one percent mass decrease and 200 million year round trip?

Hi sirxerp,
ignoring cosmological expansion the photon will red shift on its way out and blue shift on the way back so that it has exactly the same wavelength when it returns as when it it left. I am not sure what you are getting at here unless you are enquiring about the mathematical equations for what happens when cosmological expansion of the universe, during the travel time, is taken into account? In that case the photon will return redshifted.

 

[sirzerp]

Hi sirxerp,
ignoring cosmological expansion the photon will red shift on its way out and blue shift on the way back so that it has exactly the same wavelength when it returns as when it it left.

Sir, you are assuming the gravity well has the same mass the whole time.

Do you think the sun has the same mass as it did yesterday?

How about a 200 million years ago?

A returning photon from the past would be affected by a slightly a different mass, hence a different blue shift as it returned, correct?

 

[yuiop]

Sir, you are assuming the gravity well has the same mass the whole time.

Do you think the sun has the same mass as it did yesterday?

How about a 200 million years ago?

A returning photon from the past would be affected by a slightly a different mass, hence a different blue shift as it returned, correct?

OK, I see your angle now and basically agree with you. On its way out the photon would be on the surface of an expanding light sphere and the mass-energy enclosed within that expanding sphere would be constant on the way out because all the other photons leaving the star would still contribute to the gravitational mass of the star. On the way back the situation changes and the mass-energy enclosed within the radius of the returning photon will be continually decreasing and so yes, the photon should be blue shifted to a lesser extent than it was redshifted on the way out.

 

[mysearch]

Response to Kev #18

One difficulty with using the Schwarzschild solution at exactly r=R_s is that no physical observer could remain stationary at the event horizon so the equztion is not physical at that exact point.

I tend to agree, but I also think we still have a lot of gaps in our knowledge.
I have included links to some Internet articles that may be of interest:
http://www.physorg.com/news101560368.html
http://arstechnica.com/news.ars/post...s-paradox.html

The speculation of some of these articles leads to the idea that matter cannot exist at the event horizon and reduces to a more fundamental quantum/wave structure. I previously raised some questions about the implication of the Schwarzschild metric in another in another thread, although I am not sure that any satisfactory conclusion was reached:
https://www.physicsforums.com/showthread.php?t=225250

Similarly, the two observers with relative velocity each measure the others clock to ticking slower than their own clock and we know that each has a valid viewpoint even thought they seem contradictory.

I have taken this quote out of sequence because I have a different perspective, which I would like to table before making any other comments. In terms of special relativity, I am not aware of any `twin paradox` that cannot be resolved. However, I believe this statement may lead to cases where an apparent relative velocity would not lead to any time dilation between the observers.

That is exactly what I stated the local observers at A, B and C would measure. Is their measurement the only valid viewpoint? I think not.

Yes, we are in agreement with respect to the local observer. However, the question of validity appears more difficult, at least, to me. As a broad generalisation, I don’t like paradoxes. It tends to say to me that we don’t understand everything that’s going on. In part, this is why I like to look for physical interpretation that supports mathematical conjecture. However, as outlined in post #17 and the thread above, resolving the apparent disparity between what is perceived by the distant (D) and local (A) observers does appear to be problematic. I guess I can sum up my reservation by saying that I can accept the relative nature of time, but not a contradictory nature. Therefore, if (D) sees time stop in (A), I don't how you can simply run time on in (A) disregarding the implications in (D)

For example the local observers in Special Relativity cannot measure length contraction and time dilation of there own rulers and clocks but that does not mean length contraction and time dilation do not exist.

Yes, I agree, but wanted to try and clarify a previous point. If (A) measures the circumference around the black hole, which is unaffected by space expansion, and then divides by 2\pi, the implication is that the actual radius is larger due to the curvature of space at his position. So:

o If his ruler expands when orientated in the radial direction, (A) could never measure the expansion directly?

o If no, how do we define the radius [r] for any observer?

o How do you interpret the formal definition of a black hole as a region of spacetime that is not in the causal past of the infinite future?

I didn’t want to inject another derivation into this post, but it appears that you are presenting a derivation that contradicts the standard Lorentz transforms. Am I misinterpreting what you are actually saying?
http://en.wikipedia.org/wiki/Lorentz_transformation

Well, my position is that there is no single explanation. From the point of view of local observers frequency of the photon changes due to time dilation and the speed of light is constant. From the point of view of an observer at infinity (or at any constant non local location) the frequency remains constant, the wavelength is stretched as the photon rises and the speed of light increases. Both points of view are equally valid.

While we agree on the local interpretation (A), as defined by the table in #11, which is verified by experiment, I still have difficulty with your interpretation of the distant perspective (D). At one level, we only have 1 equation we are trying to balance, i.e. c=f\lambda. Your other table in #11 does this by changing [c] and \lambda, while my interpretation changes [f] based on the observed time dilation between (D) and (A, B & C). Of course, in practice, (D) cannot observe anything in (A) until photons emitted from (A) arrive at (D), which we have agreed will be redshifted to a lower frequency, but with velocity equal to [c]. I gave an example in #15, which suggested that conceptually the observed value of [c] would be [c] because the observe time to travel 1 light-second increases, but so does the radial distance. I also think the argument about the correlation between frequency-energy and wavelength-media to be more logical, but this doesn’t mean that it is.

As always, appreciate the injection of new ideas.

 

[mysearch]

General Response to #18-#21

In #19, sirzerp asked:

Anyone have the wavelength math for a one percent mass decrease and 200 million year round trip?

We might wish to start with the general shift equation:

z = \frac{GM}{rc^2}

My initial assumption is that given the outbound photon is traveling at the speed of light [c], any change in the mass/gravity in the first 100 million years would never catch up to the photon. Of course, the problem would be more complex on the way back. However, the general conclusion would seem to be that the photon would be subject to less blueshift in the second 100 million years that the initial redshift.

However, maybe this problem would be more meaningful from an energy perspective. While the inference appears to be that the photon losses energy, I believe the conservation of energy, i.e. total energy equals kinetic + potential, would be maintained. Sorry if I have missed the point of the question.

 

[sirzerp]

Delta Mass Shifts

My initial assumption is that given the outbound photon is traveling at the speed of light [c], any change in the mass/gravity in the first 100 million years would never catch up to the photon. Of course, the problem would be more complex on the way back. However, the general conclusion would seem to be that the photon would be subject to less blueshift in the second 100 million years that the initial redshift.

However, maybe this problem would be more meaningful from an energy perspective. While the inference appears to be that the photon losses energy, I believe the conservation of energy, i.e. total energy equals kinetic + potential, would be maintained. Sorry if I have missed the point of the question.

Well, let's say you live in a low mass solar system (yellow dwarf) and you move to a high mass solar system (blue giant), then the universe is going to look a little more blue-shifted that what you were used to.

If you move from a high mass solar system to a low mass solar system, then the universe is going to look a little more red-shifted than it used to.

If you live in a solar system that is gaining mass over time, then each day the universe is becoming a little more bluer each day from your viewpoint.

If you live in a solar system that is losing mass over time, then each day the universe is becoming a little more redder each day.

Here is the surprising part. If you send a active radar ping from your solar system to another solar system and if your solar system is losing mass over time, then your return pong is red-shifted!

Think of the trip, an outgoing red shift (R1) and then an incoming blue shift (B2) when it arrives at the second solar system; and then a reflection, and then an outgoing red shift (R2) and finally an incoming blue shift (B1).

First thing as the mass changes over time, notice that R1>B1 and B2>R2 so the affect should cancel right?

I don’t think so because B2 and R2 are adjacent in time. R1 and B1 are not adjacent in time. Many millions of years could have passed between when the ping when out and the pong came back.

The mass change in the R1 & B1 events is most likely larger than B2 & R2 events do the inverse square nature of gravity, and the amount of time between incoming and outgoing.

On the energy side of things, it isn’t that the photon energy is changing; it is that time-space slope has changed at your location.

It’s like riding your bike to the top of a hill, with a road crew working on the road behind you.

Your return trip has a little less acceleration because of the slope of the hill being changed during your ride.

 

[yuiop]

...

If you live in a solar system that is losing mass over time, then each day the universe is becoming a little more redder each day.

Let me guess where this is going

If we have two galaxies that are a fixed distance apart, then over time, the red shift due to loss of mass from the galaxies gives the appearance that the galaxies are moving apart and the apparent (accelerating) expansion of the universe that we observe may just be an illusion?

Without doing the calculation I guess the effect you describe is small and not sufficient to account for all the redshift we observe, but your observation is not without merit. Another difficulty is that we would require some mysterious force to counter the force of gravity in order that the galaxies remain a fixed distance apart. In fact Einstien did just that, and called his mysterious force the cosmological constant, but later withdrew it. We can not rule out a mysterious force to keep the galaxies a fixed distance apart as cosmologists have no difficulty accepting a mysterious anti-gravitational force is accelerating the expansion of the universe. Without knowing the exact magnitude of the effect you describe, I can not give a definitive answer. There are some clever people here on this forum that probably are capable of doing the calculations

 

[yuiop]

...

I didn’t want to inject another derivation into this post, but it appears that you are presenting a derivation that contradicts the standard Lorentz transforms. Am I misinterpreting what you are actually saying?
http://en.wikipedia.org/wiki/Lorentz_transformation

The derivations I presented were based on the ones shown here http://hyperphysics.phy-astr.gsu.edu/Hbase/relativ/tdil.html#c1 and here http://hyperphysics.phy-astr.gsu.edu/Hbase/relativ/ltrans.html#c2

I will come back to the rest of your post one bite at a time to avoid indegestion :P

 

[mysearch]

General Response to #25/26

We seem to have got diverted from the original question, but its throws up some interesting thoughts that are associated with one further question I had. However, to briefly respond to Sirzerp example, it might worth getting the size of the shift factor into some perspective using the equation:

z = \frac{1}{\sqrt{1-Rs/r}}-1

Please double check the calculation, but it seems that a photon falling into the surface of our sun (yellow dwarf) would only have a value of z=2.13E-6. On the assumption that our blue giant was 10x larger, it would have a value of z=2.13E-4 at its surface, because both radius are much greater than the Schwarzschild radius [Rs]. These figures are absolute shifts, whereas the example would only incur at differential shift between the two systems. Of course, the z-shift would be much less at a safe Earth orbit. So the implication is that the universe wouldn’t actually look that different, but I understand this was not the main point being raised.

However, to turn the question back to the original thread, it would seem relevant to ask what the blueshift-energy implies to an observer very close to an event horizon of a black hole? At r=3Rs, the shift would correspond to a 22% increase in [z], which I am assuming would translate to a 22% increase in frequency and energy. This would increase to 73% at the photonsphere (r=1.5Rs) and ultimately to infinity at r=Rs. However, I would question this implication and would like to hear if there is an alternative explanation.

P.S. Kev, please take your time and avoid indigestion by all means

 

[yuiop]

Hi mysearch,

I hope we are now at the point that we agree with my statement in post #14:
------------------------------------------------------------------------------------
a) Gravitational time dilation slows clocks by the gamma factor.
b) Coordinate speed of light slows by gamma squared.
c) Local speed of light is always c.

If you agree with statements a,b and c you have to come to the conclusion that length (rulers) have to contract by gamma aproaching a massive body.
------------------------------------------------------------------------------------

The following list is a comparison of kinetic versus gravitational time dilation and length contraction of a single clock and ruler:

Transverse kinetic time dilation: t' = t*y
Horizontal gravitational time dilation t' = t*g

Transverse kinetic length contraction: x' = x
Horizontal gravitational length contraction: x' = x

Parallel kinetic time dilation: t' = t*y
Vertical gravitational time dilation t' = t*g

Parallel kinetic length contraction: x' = x/y
Vertical gravitational length contraction: x' = x/g

where y and g are the usual kinetic and gravitational gamma factors and the unprimed variables are proper times and lengths of measuring devices. The very nice correspondence between gravitational and kinetic relativistic effects is obvious. Do you agree with above list?

--------------------------------------------------



In another thread of yours https://www.physicsforums.com/showthread.php?t=227410 you stated:
--------------------------------------------------------------------------------------
o As velocity [v] approaches the speed of light [c], time slows and length contracts in the direction of motion, at least, with respect to a `stationary` observer.

o On approaching a gravitational mass [M], time slows and length expands in the direction of gravitational pull, as a function of radius [r], at least, with respect to a `distant` observer.
---------------------------------------------------------------------------------------
which indicates you do not agree with my list (at that time).

Your statement would be more accurate stated as :
---------------------------------------------------------------------------------------
o As velocity [v] approaches the speed of light [c], clock slow and rulers contract in the direction of motion, at least, with respect to a `stationary` observer.

o On approaching a gravitational mass [M], clocks slow and rulers contract in the direction of gravitational pull, as a function of radius [r], at least, with respect to a `distant` observer.
---------------------------------------------------------------------------------------
Do you agree?

Jorrie and I were of the opinion that the radius calculated by a local observer by measuring circumference would agree with radius measured by the observer at infinity. The localy measured orbital velocity would be less by gamma according to the infinity observer but the difference in clock rates between the local and infinity observer ensures they get the same result for radius. This is also corresponds with there being no horizontal length contraction in my list.

... If (A) measures the circumference around the black hole, which is unaffected by space expansion, and then divides by 2\pi, the implication is that the actual radius is larger due to the curvature of space at his position. So:

o If his ruler expands when orientated in the radial direction, (A) could never measure the expansion directly?

Yes, the local observer can never measure the change in the proper length of his ruler. I prefer to think that the observer at infinity sees the ruler as length contracted when orientated vertically. The length contraction and time dilation of the local observer's rulers and clocks ensures he always measures the local speed of light to be constant (c) while the observer at infinity see the speed of falling light to be slowing by gamma squared. If the local observer tries to measure radius directly using a ruler then he would measure a greater radius than the infinity observer and he would be of the opinion that the circumference is less than 2\pi R. (Curvature of spacetime?)


This post is just to check we are "all singing from the same hymm sheet".

If not, I hope we can identify where we differ and come to some sort of agreement.

 

[mysearch]

Response to #29

Kev, just a quick reply to say that I fully occur with the sentiments of your last statement:

This post is just to check we are "all singing from the same hymm sheet".
If not, I hope we can identify where we differ and come to some sort of agreement.

I will reflect on the details of your last post and review all exchanges in the thread and attempt to summarise any points of difference. Just my way of explanation, the reason I am doing this was to see whether a consistent model of relativity would fall out of a simple model that only considered relativity in terms of the time/space effects caused by velocity and gravity. Anyway, I shall try to post a more coherent response as soon as possible, which will probably be tomorrow. Again, thanks for your time and insights.

 

[yuiop]

OK, in the meantime I am just typeing up a post about what happens below the event horizon which might be of interest to you ;)

 

[yuiop]

... However, the question of validity appears more difficult, at least, to me. As a broad generalisation, I don’t like paradoxes. It tends to say to me that we don’t understand everything that’s going on. In part, this is why I like to look for physical interpretation that supports mathematical conjecture. However, as outlined in post #17 and the thread above, resolving the apparent disparity between what is perceived by the distant (D) and local (A) observers does appear to be problematic. I guess I can sum up my reservation by saying that I can accept the relative nature of time, but not a contradictory nature. Therefore, if (D) sees time stop in (A), I don't how you can simply run time on in (A) disregarding the implications in (D)

OK, now we get onto the more interesting stuff :)

One point of view, possibly the accepted point of view, is that time runs on the same deep down in the gravitational well as for an observer at infinity and it is just the delay in information getting to the distant observer that accounts for the apparent difference in clock rates. At the event horizon the delay becomes infinite and information about what is happening there, never gets to the distant observer.

Another point of view (mine) is that the black hole never completely forms into the classic point singularity of infinite density. You could call this version the assymptotic version. In this thread https://www.physicsforums.com/showthread.php?t=223730&page=2 in post #17 onwards the interior Schwarzschild solution was discussed and it was shown that once a mass collapses to r=Rs*9/8 that time starts to reverse at the centre of the mass. The implication is that some physical process (gravity or pressure) starts to re-distribute the matter and density distribution so that all the mass ends up as thin shell just outside the Schwarzschild radius. To outward appearances an assymptotic black hole is just like a normal black hole but it allows a small amount of highly red shifted radiation to escape which gives a simpler explanation for Hawking radiation. It also makes the information loss paradox a non-paradox which agrees with the conclusion reached by the scientists in the link you posted. http://www.physorg.com/news101560368.html

The interior solution applies for coordinates below the physical surface of the gravitational mass and takes the reduced closed mass below the surface into account. The exterior Schwarzschild solution only applies to regions outside the surface of physical mass. For a classical black hole with all the mass located at the cental singularity the exterior Schwarzschild solution is valid below the even horizon all the way down to close to the singularity. The only trouble is that the exterior solution appears to yield imaginary answers below the event horizon due to taking the roots of negative numbers. There is however another way to use the exterior solution that might circumvent this problem to a certain extent and allow us to peek below the horizon. The normal gravitational gamma factor is given as:

(Eq1) \frac{1}{\left(1-Rs/R\right) }

This can be re-expressed as

(Eq2) \frac{\sqrt{1-Rs/R_o} }{\sqrt{1-Rs/R}}

Where Rs is the Schwarzschild radius, R is the radius where the event being measured is located and Ro is the radius that the observer making the measurement is located at. It is easy to see that if the observer is located at infinity Eq2 reduces to the usual Eq1.

Now we know that normally an observer sees clocks below him running slower than his own clock and objects tend to fall from locations of low time dilation to locations of high time dilation. If the observer is below the event horizon (say at R = Rs*9/10) and he is observing a clock below him at R = Rs*1/2 then the time dilation factor would be:

\frac{\sqrt{1-10/9} }{\sqrt{1-2/1}} = \frac{\sqrt{-1/9} }{\sqrt{-1}} = \frac{ (1/3) i}{i } = 1/3

where i is the square root of (-1)

It can be seen in this example the clock below the observer is running faster than his own clock suggesting a gravitational potential that is in the opposite direction to the gravitational slope outside the event horizon. From this point of view is it possible that everything falls towards the event horizon whether below or above the event horizon? Of course this solution is based on the exterior solution that assumed a point singularity which we are saying never formed. The interior solution on the other hand suggests that as the black hole forms mass is redistributed before the event horizon forms so that all the mass is in a thin shell just outside the horizon and assymptotically and infinitely slowly falling towards the horizon. The advantage of the interior solution is that we can track the distribution of matter before the singularity forms. The assymptotic interpretation solves the problem of time stopping at the event horizon according to a distant observer while progressing normally according to a local observer because there is never any matter or observers at the event horizon. All the above is meant to be within the context of General Relativity and not meant to be a disproof of that theory. It can also be noted that Einstien himself did not agree that General Relativity predicted black holes with singularities of infinite density at the centre.

 

[mysearch]

General Response to #32

Thanks for the outline of the possible internal workings of a black hole. It is something that I would like to pursue further once I have my understanding of the external workings on a firmer footing So many thanks for the links as I had not come across the details of the asymptotic proposal before.

However, I would say that I am somewhat sceptical, at this stage, that science has an adequate and coherent quantum model of matter to say what is really happening at/or below the event horizon. Main reply to #29 to follow.

 

[yuiop]

Thanks for the outline of the possible internal workings of a black hole. It is something that I would like to pursue further once I have my understanding of the external workings on a firmer footing So many thanks for the links as I had not come across the details of the asymptotic proposal before.

However, I would say that I am somewhat sceptical, at this stage, that science has an adequate and coherent quantum model of matter to say what is really happening at/or below the event horizon. Main reply to #29 to follow.

I guess we will also have to figure out the meaning and significance of tortoise coordinates and why they seem to to contradict the asymptotic model.

 

[mysearch]

Response to #29: Part 1 of 2

Kev, I have tried to respond to each section in #29 in-turn. However, apologises for the excessive length, i.e. parts 1 & 2, but I wanted to try to explain the outstanding issues based on my current understanding.

--------------------------------------------------------------
Section-1:
This section relates to bullets a, b, c as per post #14:
Yes, I agree. As per my post #17, the Schwarzschild metric leads to the equation:

\frac{dr}{dt} = \pm c/\gamma = \pm c \left(1-Rs/r\right)

\gamma = \frac {1}{\sqrt{\left(1-Rs/r\right)}}

This also confirms your table in #11. Of course, the contradiction cited in #17 still remains and questions both the meaning and validity of the equation, because it suggests that the coordinate speed of light [c] is zero at the event horizon.
--------------------------------------------------------------

Section-2:
Summarises kinetic & gravitational effects. Took me a while to orientate myself to your notation. However, if I have interpreted the notation correct, there are two that I would like to try and clarify:

Parallel kinetic length contraction: x' = x/y
Vertical gravitational length contraction: x' = x/g

The implication is that velocity and gravity both lead to length contraction, which is something I questioned. However, I wanted to be clear on the meaning of [x,x`]. The normal Lorentz transform as described in many reference, e.g. Wikipedia: http://en.wikipedia.org/wiki/Lorentz_transformation

x' = \frac{x-vt}{\sqrt{1-v^2/c^2}}

In the reduced form, where t=0, the `parallel kinetic length contraction` would appear to go to:

x' = \gamma(x) and not x' = x/y

My interpretation of this equation is as follows: [x`] is the length measure onboard the craft traveling at [v], e.g. 1 metre. While [x] is the equivalent length measured by the stationary observer. [x` > x] as the observed length is contracted due to velocity. So while we are saying the same thing in words, we appear to have different equations.

I suspect there is also a different interpretation associated with the `vertical gravitational length contraction`. It is my understanding, although this is not an assertion of fact, that the radial length expands under gravity due to the increased curvature of space as a larger gravitational mass [M] is approached. However, I would like to pursue the clarification of this point by stepping to the next section in part-2.

End of Part 1:

 

[mysearch]

Response to #29: Part 2 of 2

Section-3:
It is probably easier to replicate the 2 definitions you modified:

o As velocity [v] approaches the speed of light [c], time slows and length contracts in the direction of motion, at least, with respect to a `stationary` observer.

You change the words `time` and `length` to `clocks` and `rulers`, therefore I wanted to clarify whether there was an important physical implication in this change rather than semantics. In cosmological expansion, the overall volume of space expands, but atoms are unaffected by this expansion, i.e. atoms do not get bigger as cosmological space expands. However, I have always interpreted length/ruler contraction in a different way, i.e. conceptually the stationary observer would perceive everything in the moving frame of reference to contract in the direction of motion, even atoms. This is why [x`] is measured locally by a ruler as 1 metre, while the stationary observer perceives both the ruler and object being measured to both be contracted.

o On approaching a gravitational mass [M], time slows and length expands in the direction of gravitational pull, as a function of radius [r], at least, with respect to a `distant` observer.

Again, we have a similar change, but the term ` length expands` is changed to `rulers contract`. Based on similar assumption as outlined previously, I assumed that the local observer’s ruler would expand in-line with any expansion in space and as such, the local observer would not perceive any stretching of the ruler as its was orientated along the radial direction. However, a distant observer would perceive both the ruler and object being measured to both be expanded.

As such, this still seems to be an open issue that I would like to clarify.

--------------------------------------------------------------

Section-4:

Jorrie and I were of the opinion that the radius calculated by a local observer by measuring circumference would agree with radius measured by the observer at infinity.

Yes, I agree. By definition, the calculated radius or coordinate-r corresponds to that measured in flat spacetime. Wasn’t too sure about the next bit, as I would have thought we could simply say that the circumference was measured without reference to any velocity. However, I agree that horizontal/tangential circumference would not be affected by gravity. At this point, I would like to introduce two definitions, although you will probably disagree with the second:

Coordinate-r = circumference/2\pi
Spatial-r = \gamma(coordinate-r)

The implication of the second definition suggests that the radial distance expands, although it is probably not measurable by the local observer for the following reasons starting with your words:

Yes, the local observer can never measure the change in the proper length of his ruler. I prefer to think that the observer at infinity sees the ruler as length contracted when orientated vertically.

While you begin with `Yes`, I don’t think you will agree with my statements at the end of section-3. I was assuming space expands in the radial direction due to the increasing curvature of space. However, I was also assuming that the local ruler would also be affected and therefore the increase in the spatial-radius is not measurable locally, only observed at a distance. So, to conclude, let me try to apply my assumptions to the 2 observers under consideration, i.e. the distant observer (x, t) and the shell observer (x`, t`) within the gravity well using my assumptions:

x` > x, i.e. x’ = \gamma(x)

t` < t, i.e. t&#039;= (t)/ \gamma

The velocity [v] is a function of distance over time:

v' = x'/t' = \gamma(x)/ (t)/ \gamma = \gamma^2 (x/t)

As such, [v’] is greater than [v] by \gamma^2 as required by the very first equation. Again, apologises for such a long posting, but as stated, I wanted to try and address all the open issue from my current understanding.

 

[yuiop]

--------------------------------------------------------------

Section-2:
Summarises kinetic & gravitational effects. Took me a while to orientate myself to your notation. However, if I have interpreted the notation correct, there are two that I would like to try and clarify:

Parallel kinetic length contraction: x' = x/y
Vertical gravitational length contraction: x' = x/g

The implication is that velocity and gravity both lead to length contraction, which is something I questioned. However, I wanted to be clear on the meaning of [x,x`]. The normal Lorentz transform as described in many reference, e.g. Wikipedia: http://en.wikipedia.org/wiki/Lorentz_transformation

x&#039; = \frac{x-vt}{\sqrt{1-v^2/c^2}}

In the reduced form, where t=0, the `parallel kinetic length contraction` would appear to go to:

x&#039; = \gamma(x) and not x' = x/y

My interpretation of this equation is as follows: [x`] is the length measure onboard the craft traveling at [v], e.g. 1 metre. While [x] is the equivalent length measured by the stationary observer. [x` > x] as the observed length is contracted due to velocity. So while we are saying the same thing in words, we appear to have different equations.

I think we are odds with the notation. It might be clearer to introduce the term "proper length" which has a clear meaning and re-write my statement:

Parallel kinetic length contraction: L' = Lo/y = (X2' -X1') = (Xo2-Xo1)/y
Vertical gravitational length contraction: L' = Lo/y (X2' -X1') = (Xo2-Xo1)/g

where Lo is the proper length measured by the local observer at rest with rod/ruler.

You seem to have x' as the proper length which might be where we differ.

Refering to the wikipedia reference http://en.wikipedia.org/wiki/Lorentz_transformation

they have t&#039; = \frac{t-vx/c^2}{\sqrt{1-v^2/c^2}}

it should be clear that the primed time coordiante is not the proper time. For example at v =0.8c a proper time interval of 1 second in the rest frame of 1 second is 1.6666 seconds in the primed frame which is moving relative to the clock. Similarly the primed x' coordinate is not a proper distance coordinate.

We also know that at v=0.8c a measurement of 1 light second for the proper length Lo of a rod in the rest frame of the rod transforms to a measurement of 0.6 lightseconds in the primed frame. Time transforms to a larger value (dilation) while length transforms to a smaller value (contraction).

For time t&#039; = \frac{t-vx/c^2}{\sqrt{1-v^2/c^2}}

T&#039; = (t2&#039; -t1&#039;) = \frac{(t2-t1)-v(x2-x1)/c^2}{\sqrt{1-v^2/c^2}}

For x2=x1=0

T&#039; = \frac{(t2-t1)}{\sqrt{1-v^2/c^2}} = To*\gamma



For length x&#039; = \frac{x-vt}{\sqrt{1-v^2/c^2}}

L&#039; = (x2&#039; -x1&#039;) = \frac{(x2-x1)-v(t2-t1)/c^2}{\sqrt{1-v^2/c^2}}

Now to measure length in the primed frame we cannot simply say t2-t1=0 because the times t1 and t2 are simultaneous in the rest frame but not in the primed frame. So we have to use the reverse transformation: See post #4 of this thread --> https://www.physicsforums.com/showthread.php?t=235322 for the calculation.

The end result is:

L&#039; = (x2&#039; -x1&#039;) = (x2-x1)*\sqrt{1-v^2/c^2} = Lo/\gamma

I suspect there is also a different interpretation associated with the `vertical gravitational length contraction`. It is my understanding, although this is not an assertion of fact, that the radial length expands under gravity due to the increased curvature of space as a larger gravitational mass [M] is approached. However, I would like to pursue the clarification of this point by stepping to the next section in part-2.

End of Part 1:

Say an observer measures the radius of the orbit to be R by measuring the circumference with rulers or timing an orbital period. When the local observer tries to measure the radius directly along the radius using rulers that are length contracted according to the infinity observer, he will measure a greater radial distance than the distance R which both the local and infinity observer agree on. As you can see in this case length contracted rulers corresponds to expanded distance and this may be another source of confusion. It should be noted that the contraction of the rulers is a measurement according the infinity observer while the expansion of the radial distance is the direct measurement of the local observer.

This also confirms your table in #11. Of course, the contradiction cited in #17 still remains and questions both the meaning and validity of the equation, because it suggests that the coordinate speed of light [c] is zero at the event horizon.

In the asymptotic proposal this contradiction is resolved because actually reaching the event horizon is unatainable just as as reaching the speed of light is unatainable.

 

[yuiop]

Section-3:
It is probably easier to replicate the 2 definitions you modified:

You change the words `time` and `length` to `clocks` and `rulers`, therefore I wanted to clarify whether there was an important physical implication in this change rather than semantics. In cosmological expansion, the overall volume of space expands, but atoms are unaffected by this expansion, i.e. atoms do not get bigger as cosmological space expands. However, I have always interpreted length/ruler contraction in a different way, i.e. conceptually the stationary observer would perceive everything in the moving frame of reference to contract in the direction of motion, even atoms. This is why [x`] is measured locally by a ruler as 1 metre, while the stationary observer perceives both the ruler and object being measured to both be contracted.

I changed the word "length" to "rulers" to emphasize that I was talking about a physical object possibly of infinitesimal size, where as you seem to be using the word "length" to mean "distance". In the Bell's spaceship paradox the distance between the two accelerating rockets remains constant in the unaccelerated reference frame while it expands in the reference frame of one of the rockets. Rulers oboard one the rockets appear to contract in the unaccelerated frame while remaining constant in the the frame of one of the accelerating spaceships. So in the accelerated frame Length (rulers) contract and distance expands. The subtle difference between length and distance is the root of the that paradox that confuses many people. Physical objects length contract according to an observer with relative motion with respect to the object but the same is not true for distances.

Also, as I said before you are interpreting the primed x' to mean a local proper measurement of an object, when the primed variables usually refer to measurements made in a frame with motion relative to the rest frame of the object. However, I am not sure if there is any hard and fast rule or convention on this, but we should be consistent.

Again, we have a similar change, but the term ` length expands` is changed to `rulers contract`. Based on similar assumption as outlined previously, I assumed that the local observer’s ruler would expand in-line with any expansion in space and as such, the local observer would not perceive any stretching of the ruler as its was orientated along the radial direction. However, a distant observer would perceive both the ruler and object being measured to both be expanded.

As such, this still seems to be an open issue that I would like to clarify.

Same comments as above.

Section-4:

Yes, I agree. By definition, the calculated radius or coordinate-r corresponds to that measured in flat spacetime. Wasn’t too sure about the next bit, as I would have thought we could simply say that the circumference was measured without reference to any velocity. However, I agree that horizontal/tangential circumference would not be affected by gravity. At this point, I would like to introduce two definitions, although you will probably disagree with the second:

Coordinate-r = circumference/2\pi
Spatial-r = \gamma(coordinate-r)

I disagree with the second for a different reason than you probably imagine. Local rulers length contract according to Lo/gamma as measured by the observer at infinity. If by R_spatial you mean the distance measured from the centre of the body to the orbital radius under consideration then it is greater than R_coordinate but not by gamma. The rulers used to measure the radial distance will be length contracted to different degrees depending on how deep they are within the gravity well so the total distance would have to arrived at by integrating the total of all the individual infinitesimal rulers and the result will be much greater than simply R_coordinate*gamma. Close to an event horizon you would need an amount of local rulers than tends towards infinite to measure distances.

The implication of the second definition suggests that the radial distance expands, although it is probably not measurable by the local observer for the following reasons starting with your words...
I was assuming space expands in the radial direction due to the increasing curvature of space. However, I was also assuming that the local ruler would also be affected and therefore the increase in the spatial-radius is not measurable locally, only observed at a distance.

Well, I agree local radial distance expands but I will add that that local observers can measure it in the sense that they measure a greater radial distance than the infinity observer when they use their local rulers. They can not however, measure the length contraction of their own local rulers or rods. I hope we can agree on the difference between length (rulers) and distance as I have defined it. In the gravitational case distance expands and length is constant according to local observers while distance is constant and length contracts according to a distant observer.

 

[mysearch]

Initial Response to #37 & #38

Kev: As always you have responded with some very convincing arguments.

The subtle difference between length and distance is the root of the paradox that confuses many people.

I believe this may also be the root of my confusions, so I want to take some time to reflect on the all the issues you have raised with respect to proper length and distance. I also need to think about the `Bell's spaceship paradox` that has really got me scratching my head.

I really appreciate the help you have extended and will respond on any open issues, if any, as soon as possible. Thanks.

 

[mysearch]

Response to #37: Lorentz Transforms

I thought it might be easier to try and close specific issues one by one rather massing all the issues together, especially as the Bell's spaceship paradox has me doubting everything at the moment.

As you may have realized, I have been trying to self-learning about relativity over the last couple of months, mainly from Internet sources; so have simply adopted the notation I commonly saw in use, e.g.
http://en.wikipedia.org/wiki/Lorentz_transformation
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/ltrans.html

Both these sources show the two main transforms as follows:

x&#039; = \frac{x-vt}{\sqrt{1-v^2/c^2}} = \gamma(x-vt)

t&#039; = \frac{t-vx/c^2}{\sqrt{1-v^2/c^2}} = \gamma(t-vx/c^2)

This suggested to me that [x’>x] with [x’] being a length measured by the moving observer and [x] the equivalent length measured by the stationary observer. As such, [x] is contracted with respect to [x’].

You seem to have x' as the proper length which might be where we differ.

Wasn’t aware that I was making any inference to proper length or proper time, simply trying to follow what I thought was accepted notation. However, by the same arguments [t’>t], but to be consistent, [t’] would also have to be a measure of time as perceived by the moving observer and [t] an equivalent measure with respect to the stationary observer. I interpreted this to mean that the tick of the clock in the moving frame would be perceived to be longer than in the stationary observer. If we use your example:

v=0.8c and \gamma=1.6666
t’=1.6666 and t=1
t’ > t

Therefore, I understood this to infer that a clock in the moving frame of reference runs slower than one in the stationary frame. I realize this interpretation possibly creates a distinction between the rate of time, i.e. the tick of the clock, and the duration or the proper time interval. However, it allows [x’, t’] to both be simply viewed as quantities associated with the moving frame, i.e. no inference to proper length or proper time. Therefore, it appears we were using different notation, which then led to different interpretations. Using your notation:

L’ = Lo/y so that Lo = yL’ it implies Lo > L’

This appears equivalent to the previous notation x&#039;=\gamma(x) and, if so, we appear to have general agreement on the Lorentz transform regarding length. However, at first glance, there appears to be a difficulty with the time transform:

T’ = To*y or To = T’/y which implies To < T’

This was why I initially raised the query against your form, but you subsequently highlighted the point that possibly clarifies the situation:

It should be clear that the primed time coordinate is not the proper time

As discussed above, this is probably true, but I never put this inference on [t’]. Therefore, I wanted to check whether my explanation of [t’] resolves the different perspectives of both styles of notation? If so, have we now converged to an agreement or am I still missing any of the points you were making?

 

[yuiop]

...

As you may have realized, I have been trying to self-learning about relativity over the last couple of months, mainly from Internet sources; so have simply adopted the notation I commonly saw in use, e.g.
http://en.wikipedia.org/wiki/Lorentz_transformation
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/ltrans.html

Both these sources show the two main transforms as follows:

x&#039; = \frac{x-vt}{\sqrt{1-v^2/c^2}} = \gamma(x-vt)

t&#039; = \frac{t-vx/c^2}{\sqrt{1-v^2/c^2}} = \gamma(t-vx/c^2)

This suggested to me that [x’>x] with [x’] being a length measured by the moving observer and [x] the equivalent length measured by the stationary observer. As such, [x] is contracted with respect to [x’].

Hi mysearch,
this is a self learning curve for me too and I find these discussions with you a welcome oportunity to clear up the concepts for myself too. Your comments make me realize I do not have some ideas as rigorously defined as I imagined I had, so bear with me if I adapt as I go along and sometimes appear to contradict things I said earlier ;) First of all the use of the word "length" in your comment "[x’] being a length measured by the moving observer and [x] the equivalent length measured by the stationary observer. As such, [x] is contracted with respect to [x’]" might be a source of confusion because x' is a coordinate and not a length. [x,t] or even [x',t] are the coordinates of a momentary event and that event does not have to belong to either the S or S' reference frame. It is difficult to define if an observer is at rest with or moving with respect to a single instantaeous event so it was perhaps misleading of me to claim that x or x' is proper measurement. Similarly, x can not contract with respect to x' because it a point coordinate with no width. We can only talk of length or length contraction with reference to a pair of coordinates such as (x2-x1) to define a spatial interval and we can only talk of time dilation with respect to a pair of time coordinates such as (t2-t1) to define a temporal interval. Only intervals can have a clearly defined rest frame. A single coordinate such as [x.t] is a point in space and time and can not length contract or time dilate. The quantities L and T that I used for length contraction and time dilation are intervals and a different animal from coordinates x and t.

...
I interpreted this to mean that the tick of the clock in the moving frame would be perceived to be longer than in the stationary observer.

Your sentence leaves a lot of room for misunderstanding. If you meant "the tick (interval?) of the (moving?) clock (as measured?) in the moving frame would be perceived (by the observer co-moving with the clock?) to be longer than (the interval measured?) in the (frame of the) stationary observer." then that is not correct. The interval measured by the stationary observer (that the clock is moving relative to) will be longer than the interval measured by the observer co-moving with the clock.

...
Therefore, I understood this to infer that a clock in the moving frame of reference runs slower than one in the stationary frame.

This is true, but it does not coincide with your above statement.

...
I realize this interpretation possibly creates a distinction between the rate of time, i.e. the tick of the clock, and the duration or the proper time interval.

It's an important distinction to make. If by "tick of the clock" you meant clock rate then you should have used slower or faster to compare measurements rather than longer or shorter.
The stationary observer considers the clocks moving relative to him as running slower than his own clocks. The interval measured between two events by his own clocks can be longer or shorter than the interval measured by an observer moving relative to him. For example say we have a rocket moving relative to our observer (Anne). A flash at the centre of the rocket is observed to take one mllisecond to reach both ends of the rocket by an observer (Bob) onboard the rocket. Anne will measure a much greater interval than one millisecond for the flash at the centre to catch up with the front of the rocket and a much shorter interval than one milisecond for the flash to go from the centre of the rocket to the rear. As you can see it is not simply a case of saying an interval in Bob's frame will be measured as longer in Anne's frame.

...
Therefore, it appears we were using different notation, which then led to different interpretations. Using your notation:

L’ = Lo/y so that Lo = yL’ it implies Lo > L’

...
T’ = To*y or To = T’/y which implies To < T’

The above two statements are true with the understanding that:

Lo is a measurement of the proper length between two spatially separated markers that are at rest with the observer making the measurement and L' is the measurement of the distance between the two markers by an observer moving relative to the markers by comparing the simultaneous spatial coordinates of the two markers.

To is the interval between two events at the same location by an observer at rest with the spatial location of the two events and T' is the interval between the same two events made by an observer moving relative to the two events. The observer moving relative to the two events will not consider the events to have occurred in the same place.

L’ = Lo/y and T’ = To*y are what is commonly meant by the expressions "length contraction" and "time dilation". Dilation is alternative word for expansion and indicates the inverse nature of length contraction and time dilation.

Are we any closer to consensus yet?

 

[dude222]

The gravitational redshift is entirely due to a change in the speed of light with locations in the gravitational field.

 

[mysearch]

Response to #41:

Hi Kev,
Sorry, I haven’t had much time this weekend to really work on all the issues you raised in #37/#38, especially the Bell’s paradox. I have found a paper by 2 Japanese guys that looks very interesting but haven’t had time to read it properly, as yet. Like you, I find it very useful to have my current understanding challenged, because often I find that I have made initial assumptions that appear logical, but cannot always be defended. In this respect, I try to remember the quote:

For every problem, there exists a simple and elegant solution, which is absolutely wrong. J. Wagoner

However, I wanted to try to move along the discussion regarding the Lorentz transforms, if possible, because it seems to be an important foundation.

First of all the use of the word "length" in your comment "[x’] being a length measured by the moving observer and [x] the equivalent length measured by the stationary observer. As such, [x] is contracted with respect to [x’]" might be a source of confusion because x' is a coordinate and not a length.

I did mean length when talking about [x, x`], but you are right to highlight this issue, as it is not rigorous. I have simply assumed the following:

x = x1-x0 and x’ = x1’-x0’ such that:

x&#039;=x1&#039;-x0&#039;=\gamma((x1-x0)-v(t1-t0) ) = \gamma(x-vt)

In this respect, [x1, x0] are the coordinate offsets in a given frame of reference, while [x] is the measured separation or length between these coordinates. In this respect, your adoption of [L & Lo] is probably more correct, but I think we are trying to say the same thing.

Your sentence leaves a lot of room for misunderstanding. If you meant "the tick (interval?) of the (moving?) clock (as measured?) in the moving frame would be perceived (by the observer co-moving with the clock?) to be longer than (the interval measured?) in the (frame of the) stationary observer." then that is not correct.

Apologises for the confusion. I was simply trying to rationalise what might appear to be a contradiction between the rate of time and the duration. The rate that a clock runs in the stationary frame, e.g. [r], is faster than the moving frame, e.g. [r’], such that we might say [r > r’]. This is what I meant by the ticking of the clock. For example, in both frames, each observer perceives time passing at the rate of 1 second per second. However, if we assume \gamma=2, there will be 2 ticks of the clock in the stationary frame for 1 tick in the moving frame, hence r > r’, meaning that the stationary observer perceives the rate of time in the moving frame to be ½ second per second. However, the reverse of this logic means that 2 seconds [t’] in the moving frame is only measured as 1 second in the stationary frame [t], hence t’ > t. This was my interpretation of the orientation suggested by the Lorentz transform

t&#039;= \gamma(t-vx/c^2) = \gamma(t) when x=0

I hope this is enough to clarify what I was trying to say, as overall, I believe we are essentially saying the same thing regarding the Lorentz transforms, but have adopted different notation to express the concepts stemming from them, although you might still disagree. At this stage, I have deliberately avoided using proper distance and proper time, as I still need to respond to the many doubts you raised in my mind about such concepts, aka the Bell paradox, along with all the equivalent effects under gravity. Hopefully, I will have some more time tomorrow.

 

[mysearch]

Response to #42:

Sorry, I am not sure that I fully understood the implications of your statement. Within all local frames, the speed of light is [c]. As such, the gravitational redshift might be interpreted as either an energy loss to the gravitational field (Newtonian) or a frequency shift due to time dilation (relativity), which in-turn affects energy by virtue of E=hf. Is there another accepted explanation?

 

[yuiop]

Hi Kev,
Sorry, I haven’t had much time this weekend to really work on all the issues you raised in #37/#38, especially the Bell’s paradox. I have found a paper by 2 Japanese guys that looks very interesting but haven’t had time to read it properly, as yet.

I posted somthing about Bell's spaceship paradox here (post#121)https://www.physicsforums.com/showthread.php?t=210634&page=9 that I hope you will find useful.

I love your quote of J Wagoner "For every problem, there exists a simple and elegant solution, which is absolutely wrong." I got a good chuckle from that. Might steal it and use it as a sig ;)

However, I wanted to try to move along the discussion regarding the Lorentz transforms, if possible, because it seems to be an important foundation.

I did mean length when talking about [x, x`], but you are right to highlight this issue, as it is not rigorous. I have simply assumed the following:

x = x1-x0 and x’ = x1’-x0’ such that:

x&#039;=x1&#039;-x0&#039;=\gamma((x1-x0)-v(t1-t0) ) = \gamma(x-vt)

In this respect, [x1, x0] are the coordinate offsets in a given frame of reference, while [x] is the measured separation or length between these coordinates. In this respect, your adoption of [L & Lo] is probably more correct, but I think we are trying to say the same thing.

The moving observer using the equation x&#039;=x1&#039;-x0&#039;=\gamma((x1-x0)-v(t1-t0) ) = \gamma(x-vt) and using t1-t0=0 will arrive at the conclusion that objects with relative motion length expand. The problem with that conclusion is the assumption that when t1 and t0 are simultaneous in S that they are simultaneous in S' too which is not correct. To correct the equation so that the measurements of both ends of the moving rod are made simultaneously then t0+(x1-x0)*v/c^2 should be substituted for t1 to get

x&#039;=x1&#039;-x0&#039;=\gamma((x1-x0)-v(t0+(x1-x0)*v/c^2-t0) )

x&#039;=x1&#039;-x0&#039;=\gamma((x1-x0)-v((x1-x0)*v/c^2) )

x&#039;=x1&#039;-x0&#039;=((x1-x0)-v((x1-x0)*v/c^2) )/\sqrt{1-v^2/c^2}

x&#039;=x1&#039;-x0&#039;=((x1-x0)(1-v^2/c^2) )/\sqrt{1-v^2/c^2}

x&#039;=x1&#039;-x0&#039;=(x1-x0)\sqrt{1-v^2/c^2}

x&#039;=x1&#039;-x0&#039;=(x1-x0)/\gamma

which is a more meaningful useful equation because the measurement is made when t1'=t0'.

The corresponding time transofrmation of an interval is:

t&#039;=t1&#039;-t0&#039;=\gamma((t1-t0)-v(x1-x0)/c^2)

t&#039;=t1&#039;-t0&#039;=(t1-t0)*\gamma

when x1=x2

 

[yuiop]

Hi mysearch,

I have an alternative derivation of x &#039;=x1&#039;-x0&#039;=(x1-x0)/\gamma using the invariant interval that might help convince you.

Starting with

(1) t=(t&#039;+vx&#039;/c^2)\gamma

Substitute the Lorentz transformation for x' in terms of x in (1) [this will be helpful later].

(2) t=(t&#039;+v((x-vt)\gamma)/c^2)\gamma which simplifies to

(3) t= \gamma t&#039; + vx/c^2

Now using the invariant interval S^2 = \Delta x^2-c^2\Delta t^2 =\Delta x&#039;^2-c^2\Delta t&#039;^2

(4) (x1&#039;-x0&#039;)^2-(ct1&#039;-ct0&#039;)^2 = (x1-x0)^2-c^2(t1-t0)^2

Since we require (x1'-x0') when t1' = t0'

(5) (x1&#039;-x0&#039;)^2= (x1-x0)^2-c^2(t1-t0)^2

Substitute the expression for t in terms of t' obtained from eq (3)

(6) (x1&#039;-x0&#039;)^2= (x1-x0)^2-c^2(\gamma t1&#039; + vx1/c^2-(\gamma t0&#039; + vx0/c^2)^2)

(7) (x1&#039;-x0&#039;)^2= (x1-x0)^2-c^2(\gamma (t1&#039;-t0&#039;) + v/c^2(x1-x0))^2

Since it has been required that t1'=t0'

(8) (x1&#039;-x0&#039;)^2= (x1-x0)^2-c^2(v/c^2(x1-x0))^2

(9) (x1&#039;-x0&#039;)=(x1-x0)\sqrt{1-v^2/c^2}=(x1-x0)/\gamma

 

[mysearch]

Response to #45/46

Hi Kev,
Glad you liked the quote, its a good reminder and thanks for the input. However, I thought we might try to anchor our respective derivations in a practical example because I am still not sure whether we are interpreting the notation in different ways, but meaning the same thing? Let’s start with the reference you provided to the Lorentz Transformation, as I have been quoting these all along, and then construct an example based on the most basic postulate of special relativity, i.e. the invariance of the speed of light in all inertial frames, i.e.

c = x/t = x’/t’

In the example, (A) will be the stationary frame and (B) the moving frame. The velocity of (B) with respect to (A) is 0.866c, giving a Lorentz factor \gamma=2. Let’s assume both frames are synchronised at the spacetime coordinates [x0, t0] and [x0’, t0’], but verify this assumption via the Lorentz transforms:

x0^{\prime} = \gamma (x0-vt0) = 2(0-(.866c)0) = 0

t0^{\prime} = \gamma ( t0-vx0/c^2) = 2(0-(.866c)*0/c^2) = 0

At this point of synchronisation, observers in (A) and (B) see a pulse of light fired at a target, which is stationary with respect to (A) and 1 lightsecond away. Now both must see the pulse of light moving away at [c] and we might readily calculate that (A) sees the pulse hit the target after 1 second, but what happens in (B)? Just for reference, light covers 2.99E8 metres in 1 second and 0.866c = 2.59E8m/s:

x1^{\prime} = \gamma (x1-vt1) = 2(2.99E8-(2.59E8)1) = 0.8E8 metres

t1^{\prime} = \gamma ( t1-vx1/c^2) = 2(1-(2.59E8)*2.99E8/c^2) = 0.267 seconds

At first, the Lorentz transforms appear to have given an anomalous answer as it suggests that the event that occurred after 1s in (A) occurs after only 0.267s in (B). However, I believe this corresponds to the `line of simultaneity` between the 2 frames, although this is more easily understood via the attached spacetime diagram. Even so, we might wish to confirm the velocity of light [c] as determined by the observer in (B) to check for a mistake, i.e. 0.8E8/0.267 = 2.99E8m/s. As such, we appear to have correlated the constancy of [c] via the Lorentz transforms, so how do we correlate the distance?

Of course, if (B) considers itself stationary, then the target must have a closing velocity of 0.866c with respect to (B). As such, the target starts from an offset comparative to [x1, x1’] and moves towards the light pulse in the (B) frame. So the total distance is the distance covered by the light pulse, i.e. [0.8E8m] plus the distance covered by the target traveling at 0.866c in the same time, i.e. 0.267s, which gives us a figure of 0.692E8m. Therefore, the comparative distance in (B) to the 1 lightsecond (2.99E8) covered in (A) is (0.8E8+0.69E8=1.49E8), which then corresponds to an expected length contraction proportional to \gamma=2.

x’ = (x)/\gamma = 2.99E8/2 = 1.49E8m such that x’ < x

The time calculated in (A) corresponded to the time for the light pulse to cover 1 lightsecond, (2.99E8m). As such, a comparative time [t’] in (B) would be that needed for light to cover 1.49E8m, i.e. 1.49E8/c=0.5s. As such, we now have a value of [t’] that proportional to \gamma=2, where t` < t.

In part, I believe the confusion over whether [x`<x] or [x’>x] depends on how you interpret the orientation, although I suspect theory dictates how this should be done. In the previous example, the length in (A) equalled 2.99E8m, i.e. 1 lightsecond, which at the speed of light is covered in 1 second. In the example, the Lorentz transforms calculate the length and time measured in (A) into the comparable length and time in (B), i.e. 1.49E8m and 0.5seconds based on \gamma=2. This appears to correspond to length contraction and time running slower in (B) compared to (A). However, what this also appears to imply is that a length in (B), e.g. 2.99E8m, is contracted to 1.48E8m to an observer in (A) because of (B) relative velocity with respect to (A).

This said, if we ignore the ambiguity of the wording in the last paragraph, are the figures in the example consistent with what we have both been trying to say?

 

[yuiop]

Hi mysearch,

Your calculations, diagram and line of simultaneity seem O.K.

...

x^{\prime} = (x)/\gamma = 2.99E8/2 = 1.49E8m such that x’ < x

This is Ok but your use x' to symbolise (x1'-x0') is confusing because x' is also used to sybolise the Lorentz transform x{\prime} = \gamma (x-vt1)

Perhaps it would be clearer to use something like \Delta x&#039; to symbolise x1'-x0' or just use x1'-x0' explicitly rather than just x' to avoid confusion with x{\prime} = \gamma (x-vt1)

It is also worth noting that \Delta x&#039; &lt; \Delta x is not always true unless \Delta t&#039; =0 In the case when t'=0 I have been using the notation L' as shorthand for \Delta x&#039; when \Delta t&#039; = 0 and it is always true that L' = Lo/y where it is understood Lo is the proper measurement along the x axis.

Hi Kev,
...
The time calculated in (A) corresponded to the time for the light pulse to cover 1 lightsecond, (2.99E8m). As such, a comparative time [t’] in (B) would be that needed for light to cover 1.49E8m, i.e. 1.49E8/c=0.5s. As such, we now have a value of [t’] that proportional to \gamma=2, where t` < t.

Your calculation that the interval t1'-t0' is less than t1-t0 for a light signal to travel from one location to another in frame A (or S) is OK but it is not a comparison of clock rates because the interval t1'-t0' is made using a single clock in frame B (or S') while the same interval is measured using two spatially separated clocks in frame A. The time interval t1'-t0' is not always less than t1-t0 as can be seen by doing the calculation for a light signal going in the opposite direction from x1 to x0. What we can always say is that if a time interval in frame A is measured in frame A by a single clock in A then B will measure that time interval to be greater. The notation I used to specify the measurement of a time interval in frame A as measured by a single clock in frame A is To and that time interval will always be measured as T' = yTo in frame B. T' and To is more about the relative rate clocks run at in different frames than a comparison of interval between events

To summerise:

x^{\prime} = \gamma (x-vt)
t^{\prime} = \gamma ( t-vx/c^2)

and

L' = Lo/y
T' = To*y

Where Lo and To are measurements of a ruler or clock by an observer at rest with the ruler or clock and L' and T' are measurements made by an observer moving relative to the ruler or clcok.

From the invariant interval

\Delta x&#039;^2 -c^2\Delta t&#039;^2 = \Delta x^2 - c^2\Delta t^2

it is fairly easy to see that \Delta x&#039; &lt; \Delta x is not always true and neither is\Delta t&#039; &lt; \Delta t

 

[mysearch]

Response to #48:

I think we are beginning to converge, although I accept that some of my notation is still probably incorrect and confusing, so will try to avoid this in the future, e.g.

This is Ok but your use x' to symbolise (x1'-x0') is confusing because x' is also used to symbolise the Lorentz transform x{\prime} = \gamma (x-vt1)

If we basically agree on the example calculation and the spacetime diagram, I feel most of the other problems are mainly to do with understanding the orientation of the specific example under discussion. Therefore, given the other valid questions raised in #37/38, I would now like to return to this aspect of the discussion, because I am still unsure about the issue of space contraction/expansion in a gravitational field. However, before addressing these issues I wanted to get a better handle on the issue of the Bell's spaceship paradox. In my defence, I will cite the following Chinese saying:

One who asks a question is a fool for five minutes; one who does not ask a question remains a fool forever.

Thanks

 

[mysearch]

Response to #37/#38

This posting is trying to get clarification or verification of a number of concepts raised throughout this thread, in particular postings #37/#38.

o SpaceTime Interval
o Proper Time
o Proper Length

The attachment st1.jpg is used as an initial reference to a Minkowski or spacetime diagram. Based on this geometry, the proper time is defined as:

[1] (Proper Time)^2 = (t^2 - x^2/c^2)

Within the causal past and future, [t] is normally greater than [x], while the interpretation of proper time is given below. Of course, in the case of (e3), [x] can be greater than [t], which would require the square root of a minus number. Therefore, in this case, it seems that the alternative definition of the spacetime interval is often preferred:

[2] (Space Interval)^2= (x^2-(ct)^2)

In essence, both these quantities seem to represent the spacetime separation in units of time or distance. The following definitions of `proper time [To]` and `proper length [Lo]` may not be rigorous, but hopefully anchors some meaning to a specific example. These definitions align to the statements made in #37/48, i.e. T = To*y and L = Lo/y, but it is believed that some caution needs to be exercised in how these statement correspond to time dilation and length contraction as implied by the Lorentz transform.

Definition of Proper Time [To]
Proper time [To] might initially be defined as the `wristwatch` time of an observer moving between 2 events with velocity [v]. This time would differ from the time [T] of a stationary person watching the observer move between the 2 events such that:

[3] T = To*\gamma

See attachment st2.jpg for the equivalent spacetime diagram. The diagrams alludes to 2 events (A) and (B). A stationary observer measures the time (5) and distance (3) taken by a moving observer to reach (B) from which the proper time corresponding to the wristwatch time of the moving observer is calculated to be 4. This suggests that time for the moving observer is running slower than the stationary observer. On the basis that the velocity [v] is invariant between both the stationary and moving observers, we might also determine the distance/length between (A) and (B) as determined by both observers, i.e.

Stationary Observer: x=vt=0.6*5=3
Moving Observer: x’=0.6*4=2.4.

The implication being that the length with respect to a moving observer is contracted. However, we will see how this statement might be misleading in some examples.

Definition of Proper Length [Lo]
Proper length (Lo) might initially be defined as the length of an object observed by a person at rest with respect to the object. This length would differ from the contracted length [L] observed by a person moving with respect to the object such that:

[4] L = Lo/\gamma

Based on the previous example in attachment st2.jpg, we saw the stationary observer measure the distance between events (A) and (B) to be 3. By the definition of proper length above, the length measured by the observer stationary to (A) and (B) was the proper length, i.e. Lo=3. As such, we could calculate the contracted length [L] from equation [4] as 2.4.

However, the proper length is said to be invariant on the basis that it always represents the maximum length seen by any observer. Possibly, the easiest way to visualise this is by assuming (A-B) to be the length of the spaceship in which the moving observer travels. Initially, before launch, both observers perceive the length to be 3 units. After take-off, (A-B) are now stationary with respect to the moving observer, not the stationary observer, as per our previous example. By definition, the proper length is invariant and a maximum, e.g. 3, and therefore the contracted length is that observed by the stationary observer, e.g. 2.4

Let’s take this line of logic one step further and assume that there are 2 identical spaceships. One used by the moving observer and one left behind with the stationary observer. Now, even though there is no ambiguity about which observer is actually moving relative to the another, the definition of proper length seems to imply that the moving observer perceives his spaceship to be 3 units in length, while perceiving the length of the spaceship on the ground to be only 2.4 units. While, the stationary observer on the ground perceives the lengths to be the other way round.

Proper Time Caveat?
If the moving observer’s was accelerating between events (A) and (B), would the proper time, i.e. wristwatch time, be shorter than that measured by a non-accelerated (inertial) wristwatch?

If yes, does the equivalence principle between acceleration and gravity, allow this additional time dilation to be interpreted in terms of an equivalent gravitational effect in conjunction with velocity?

Proper Length Caveat?
If yes, is there an equal implication of there being an extra effect on space curvature due to this acceleration as per gravity?

If curvature increases with gravity, does this mean that light has to travel a greater distance than implied by the coordinate-radius?