What Happens to Light Speed When Measured from a Fast-Moving Space Station?

[Janus]

If the rocket measures it to be 300,000 km in its own frame, the station will also measure it to be 300,000 in its own frame. Each will measure it to be shorter in the other's frame, however.
Kinda goes without saying.

No, That's not right.

Let's try this again with the following attachment.

Assume two very long trains. each car is 20,000 km long, as measured from the train to which it belongs.
Train 1 has three observers, M, A and B, A and B are 300,000 km from M, according to M, which puts them 15 cars away from M.
Train 2 is moving at .5c relative to train 1. It has observers M', A' and B'.
A" and B' are each placed an equal distance from M' such that from the frame of M, A and B, when M and M' pass each other, A' passes A and B' passes B. since train 2 is length contracted according to train 1, this places A' and B' about 17 cars away from M' on train 2. This means that the distance between A' and M' (or B' and M') is 346420 km according to train 2.

Now look at what happens according to train 2. A' and B' are a little over 17 cars from M' (this has not changed) putting them 346420 km from M'
A and B are still 15 cars from M, but since train 1 is length contracted according to train 2, each of these cars is only 17320 km long, meaning that A and B are, according to train 2, only 259800 km from M.
When M passes M' A' does not pass A and B' does not pass B.

so if the rocket measures the distance as 300,000km in its frame, the station does not measure it to be 300,000km in its frame.

 

[Doc Al]

Incidently, when we multiply gamma times length and get 346,410.15 and -346,410.15, what do these numbers tell us about A and B in relation to M, other than that they are still equidistant from M?

I assume you are talking about the \gamma L that appears in these equations:

B\'\; = (\gamma L,\frac{-\gamma\beta L}{c})
A\'\; = (-\gamma L,\frac{\gamma\beta L}{c})​

(I just noticed that jcsd defined the lights opposite to the way I would. In my set up, the lights are arranged in order A, M, and B. M' moves towards B. I hope this doesn't add any confusion.)

If so, then you must understand what they mean. These are the space-time coordianates according to the train frame of the events (A) light flashes at A and (B) light flashes at B. They tell us that:

- when light A flashed it was at position x' = -\gamma L in the train frame
- when light B flashed it was at postion x' = \gamma L in the train frame​

Yes, those flashes occur at postions that are equidistant from the midpoint M' of the train (but they occur at different times).

 

[zoobyshoe]

so if the rocket measures the distance as 300,000km in its frame, the station does not measure it to be 300,000km in its frame.

Somewhere back there I gave the distance A---->B as 600,000 km. That's in the embankment frame.

I have also given the train a length of 600,000 km in its own frame wherever I mentioned a length for the train.

This means that when the train is at rest in the embankment frame they both measure 600,000 km.

When the train is moving at .5 c, it still measures itself to be 600,000 km long. The distance A--->B in the embankment frame, is still measured to be 600,000 km long in its frame.

If the train measures the embankment while moving at .5 c, it is, of course a different story. The moving train will see the distance A--->B contracted from 600,000 to .8660254% of that which is 5196152.4 km.

If an observer in the embankment frame measures the train, it will find the train to have contracted to 519,615.4 km, as well, from the embankment viewpoint.

If we need to demonstrate the same for the ship example for any reason we can do so by attaching your rods, provided the rods give it a length of 600,000 km in its own frame.

 

[zoobyshoe]

I assume you are talking about the \gamma L that appears in these equations:

B\'\; = (\gamma L,\frac{-\gamma\beta L}{c})
A\'\; = (-\gamma L,\frac{\gamma\beta L}{c})​

That's right.

(I just noticed that jcsd defined the lights opposite to the way I would. In my set up, the lights are arranged in order A, M, and B. M' moves towards B. I hope this doesn't add any confusion.)

Nope.

If so, then you must understand what they mean. These are the space-time coordianates according to the train frame of the events (A) light flashes at A and (B) light flashes at B. They tell us that:

- when light A flashed it was at position x' = -\gamma L in the train frame
- when light B flashed it was at postion x' = \gamma L in the train frame​

OK. This is exactly what I thought. However, what confuses me is that 346,410.15 km is a greater distance than 300,000 km. I would expect it to be contracted, not lengthened. Of what use is it to apply the time dilation equation to a length? What does this help us understand about the situation?

Yes, those flashes occur at postions that are equidistant from the midpoint M' of the train (but they occur at different times).

 

[Doc Al]

OK. This is exactly what I thought. However, what confuses me is that 346,410.15 km is a greater distance than 300,000 km. I would expect it to be contracted, not lengthened.

You are still thinking that \gamma L is the length L as measured by the train. It's not. According to the train, things go like this: Light B flashes. Then, some time later, light A flashes. During that time the train is moving (of course, in the train frame, the train sees the platform move).

Of what use is it to apply the time dilation equation to a length? What does this help us understand about the situation?

I'm not sure what you mean by "apply the time dilation equation" to a length, as I did no such thing. I applied the Lorentz transformation to the space-time coordinates of an event.

If you wish to think of sticks shrinking and clocks moving slowly, then you apply the LT to sticks (lengths) and clocks. For example, a length of L in the embankment is measured to be L/\gamma from the train (yes, the train measures the lights A and B to be only 2L/\gamma apart); and the train frame will measure a time of \gamma \Delta t when an embankment clock measures only a time of \Delta t.

To analyze a problem correctly you need to apply length contraction, time dilation, and the relativity of simultaneity all together. That's tricky. The Lorentz transformation takes account of all of that for you.

 

[zoobyshoe]

In the same online version of Relativity from which I've been quoting parts of chapter IX, Einstein gives the Lorentz transformation as:

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

y'=y

z'=z

t'=\frac{t-\frac{v}{c^2}•x}{\sqrt{1-\frac{v^2}{c^2}}}

The two interesting ones, of course, are those for x' and t'.

In the next chapter, Einstein gives a simpler version of x':

^x(beginning of rod)^=0\sqrt{1-\frac{v^2}{c^2}}

^x(end of rod)^1•\sqrt{1-\frac{v^2}{c^2}}

"The distance between the two points being
\sqrt{1-\frac{v^2}{c^2}}

If we want to express A or B as A' or B', it seems to me this is the one to use, multiplying it by the length, L, which is the distance between the two points (M--->B, or M--->A) in the embankment frame that we want to locate in the train frame.

Instead, jcsd used the next one referred to by Einstein; the one he uses to find the time dilation between two successive ticks of a clock. The first tick is t = 0. The second tick is:

t=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}

If the 1 in the numerator of the equation given by jcsd for \gamma is not the one used by Einstein in the time dilation version, I am quite confused about how jcsd arrived at the 1 from x-vt, which is the numerator in the more general equation for x'.

 

[zoobyshoe]

Those are from this chapter:

Chapter 11. The Lorentz Transformation. Einstein, Albert. 1920. Relativity: The Special and General Theory
Address:http://www.bartleby.com/173/11.html

And the following one:

Chapter 12. The Behaviour of Measuring-Rods and Clocks in Motion. Einstein, Albert. 1920. Relativity: The Special and General Theory
Address:http://www.bartleby.com/173/12.html

 

[Doc Al]

In the same online version of Relativity from which I've been quoting parts of chapter IX, Einstein gives the Lorentz transformation as:

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

y'=y

z'=z

t'=\frac{t-\frac{v}{c^2} x}{\sqrt{1-\frac{v^2}{c^2}}}

The two interesting ones, of course, are those for x' and t'.

Yes, those are the same Lorentz transformations that jcsd and I both used.

In the next chapter, Einstein gives a simpler version of x':

^x(beginning of rod)^=0\sqrt{1-\frac{v^2}{c^2}}

^x(end of rod)^1 \sqrt{1-\frac{v^2}{c^2}}

"The distance between the two points being
\sqrt{1-\frac{v^2}{c^2}}

Careful. This is not a "simpler version of x' ", it is a specific application to find how a length gets transformed when observed from a moving frame. Note that each end must be located at the same time.

If we want to express A or B as A' or B', it seems to me this is the one to use, multiplying it by the length, L, which is the distance between the two points (M--->B, or M--->A) in the embankment frame that we want to locate in the train frame.

If the problem is: Find out the distance from A to B as seen by the train frame, then that would be correct. But that's not what jcsd was calculating. Instead he (and I) calculated when and where those two flashes occurred according to the moving frame. That's a different problem. For one thing, those flashes occur at different times--so that length contraction formula doesn't apply.

jcsd used A to represent the coordinates of the event "Light A flashes" in the embankment frame, and A' to represent the coordinates of that same event in the train frame. (I'm not crazy about this notation, but that's what he used.)

How did we get the answers we did? It's easy. We know the coordinates of the events in the embankment frame, that's a given: x = -L, t = 0 (for A flashing) and x = +L, t = 0 (for B flashing). Now to find out when and where the train observers determine these flashes took place: apply the LT. Try this for yourself!

Instead, jcsd used the next one referred to by Einstein; the one he uses to find the time dilation between two successive ticks of a clock. The first tick is t = 0. The second tick is:

t=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}

This is not what jcsd did. This particular equation describes the time dilation of a moving clock.

 

[plover]

In the same online version of Relativity from which I've been quoting parts of chapter IX, Einstein gives the Lorentz transformation

This is a four dimensional version of the transformation I gave above. As it shows, spatial distortions happen only along the direction of motion (in this case along the x-axis). This is why the two dimensional system suggested by jcsd is useful.

Note: When Doc Al says: "jcsd defined the lights opposite to the way I would", he makes precisely the same change as I did where I said "I've switched the signs so that the train moves toward positive x_m".​

The equation I gave was:

(x', t') = \Lambda (x, t) = \left( \gamma (x - ut), \gamma(t - \frac{ux}{c^2}) \right)
​

This is entirely equivalent to the separate equations:

x' = \gamma (x - ut)

t' = \gamma \left(t - \frac{ux}{c^2} \right)
​

(I expect this goes without saying -- I'm just ensuring the intention is clear.)

The notations introduced by jcsd (and which I should perhaps have reiterated):

\beta = \frac{u}{c}
<br /> \gamma = \frac {1} { \sqrt{1 - \beta^2} } = \frac {1} { \sqrt{1 - \frac{u^2}{c^2}} }<br />​

These notations are quite standard. Using u for the velocity in SR calculations also seems to be a current standard practice.

Again just to be clear, I'll apply these definitions to the separated equations above:

<br /> x&#039;\ =\ \gamma (x - ut)<br /> \ =\ \frac{1}{\sqrt{1 - \beta^2}} (x - ut)<br /> \ =\ \frac{x - ut}{\sqrt{1 - \frac{u^2}{c^2}}}<br />

<br /> t&#039;\ =\ \gamma \left( t - \frac{ux}{c^2} \right)<br /> \ =\ \frac{1}{\sqrt{1 - \beta^2}} \left(t - \frac{ux}{c^2} \right)<br /> \ =\ \frac{\left(t - \frac{ux}{c^2} \right)}{\sqrt{1 - \frac{u^2}{c^2}}}<br />​

These are obviously identical to the equations given by Einstein.

Note: There is a typo in Ch 12 of the Einstein. Near the end of paragraph 1 a sentence reads:

"For the velocity v = 0 we should have

\sqrt{1 - v^2/c^2} = 0​

and for still greater velocities the square-root becomes imaginary."
​

This sentence only makes sense if it is changed to read: "For the velocity v = c we should have...". (Someone should probably back me up on this.)
​

There is one issue that everyone has neglected so far (I think). Einstein's coordinate system diagram (in Ch 11) shows that the k system views the k' system as moving to the right, i.e. toward positive x. From the viewpoint of k', however, k is moving to the left, i.e. toward negative x. Thus if k measures the velocity of k' to be u then k' will measure the velocity of k to be -u. Similarly, if the x axes of the train and embankment frames are oriented to be positive in the same direction, then if an embankment observer measures the velocity of the train as u, an observer on the train measures the velocity of the embankment to be -u.

If I'm reading everyone correctly, jcsd and Doc Al took this into account (though I don't think they mentioned it). I, on the other hand, um, managed to make the issue disappear by making a calculation error when I applied the inverse Lorentz transform in my previous post, which I should fix (and which is probably why zoobyshoe found my argument for jcsd's values ambiguous...).

I assume it is agreed that the coordinates in the embankment frame for the event "flash at B" (which we are also taking to be the coordinates of "front of train passes B") are:

x = L
t = 0​

If we insert these values into Einstein's equations (and reapply the standard notations), we get:

<br /> x&#039;\ =\ \frac{x - ut}{\sqrt{1 - \frac{u^2}{c^2}}}<br /> \ =\ \frac{L - u \cdot 0}{\sqrt{1 - \frac{u^2}{c^2}}}<br /> \ =\ \frac{1}{\sqrt{1 - \beta^2}}\ \cdot\ L<br /> \ =\ \gamma L<br />

<br /> t&#039;\ =\ \frac{\left(t - \frac{ux}{c^2} \right)}{\sqrt{1 - \frac{u^2}{c^2}}}<br /> \ =\ \frac{\left(0 - \frac{uL}{c^2} \right)}{\sqrt{1 - \frac{u^2}{c^2}}}<br /> \ =\ \frac{-1}{\sqrt{1 - \beta^2}}\ \cdot\ \frac{u}{c}\ \cdot\ \frac{L}{c}<br /> \ =\ \frac{-\gamma\beta L}{c}<br />​

So if we set u to .5c our value for x' is:

<br /> \beta\ =\ .5c/c\ =\ .5<br />

<br /> \gamma\ =\ \frac{1}{\sqrt{1 - (.5)^2}}\ =\ \sqrt{4/3}\ =\ 1.15<br />

<br /> x&#039;\ =\ \gamma L\ =\ 1.15L<br />​

Now since:

• the origin of the x' axis of the train frame is the center of the train, and
• the front of the train is at the point (x, t) that we transformed from the embankment frame

we know that the value of x' also equals half the length of the train. So at rest the length of the train is: 2 * 1.15 * L. Now why should this be so?

The train as measured in the embankment frame has length 2L. However, the train is in motion in the embankment frame, and thus appears contracted in that frame. So, since the value we started out with measures the contracted length of the (moving) train in the embankment frame, it makes sense that in the train frame, where the train is at rest, the length - no longer being contracted - is measured to be greater.

 

[plover]

Sophist.

Any length that one observer measures to be contracted will be its original length when "uncontracted, not greater than its original length.
_______________________________________________

he/she may simply have made that whole post up. I certainly don't trust anything he/she says farther than I can throw it. The "uncontraction" post where the length "uncontracts" to one larger than it had before has made me suspicious.

What are you talking about?

I used the word "uncontraction" to refer to the purely mathematical operation of finding the original length given a contracted length. When did I ever claim that something would be greater than its original length?

And what have I done that would lead you to countenance that I might intend such a reading? That makes a suspicion of willful deceit or "making whole posts up" more plausible than a mistake on one or both of our parts?

What's really going on here? (If you've got some beef, my pm box is open...)

 

[zoobyshoe]

Careful. This is not a "simpler version of x' ", it is a specific application to find how a length gets transformed when observed from a moving frame.

Good point. "Simpler version" is not particularly accurate. "Specific application" is much more to the point, much more accurate. I appreciate the distinction.

Note that each end must be located at the same time.

By "at the same time" do you mean at a specific time coordinate such as T=0? Or something else?

Instead he (and I) calculated when and where those two flashes occurred according to the moving frame.

This is what I have understood you to be saying about this since I raised my objection. However, I haven't been able to understand from you how the answer that is produced: -346, 410.16 and + 346,410.16, contains any useful information about the when and where according to the moving frame. If I observe that the A or B event took place in the embankment frame at t=0, x=+/-300,000 km. What does a value of -/+346,410.16 tell me about the moving frame?

According to you, this number says something useful about the when and where of event A' or B' in the moving frame. If this number actually has any use, I can't see it. In fact, to me it looks like it is saying something both incorrect and useless about these events in the A' and B' frame. These numbers seem to me to say nothing at all about the when in the A' and B' frame, and to say something outright incorrect about the position A' and B' will have in the moving frame.

From the general direction and tone of your answers, it seems clear that you understand the difficulty I am having with these numbers. I am trying to phrase my difficulty more specifically in the hope you'll see how to solve it.

jcsd used A to represent the coordinates of the event "Light A flashes" in the embankment frame, and A' to represent the coordinates of that same event in the train frame.

Yes, this is what I have understood him to be doing.

How did we get the answers we did? It's easy. We know the coordinates of the events in the embankment frame, that's a given: x = -L, t = 0 (for A flashing) and x = +L, t = 0 (for B flashing). Now to find out when and where the train observers determine these flashes took place: apply the LT. Try this for yourself!

Specify which of the four you wish me to try for myself. I can use the one for x' and I can use the one for t' but there is no one of the four that gives an answer for both x' and t'. You are giving me one equation and claiming the answer is telling me both the x' and t' coordinates with one number.

This is not what jcsd did. This particular equation describes the time dilation of a moving clock.

jcsd gave:

\gamma=\frac {1}{\sqrt{1-\beta^2}}

where:

\beta = \frac{u}{c}

This is exactly the same equation as:

t=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}

solving for u/c separately and then squaring the result
\beta^2

gives the same answer as: the square root of u²/c².

I need for you and jcsd to justify giving me the special application of the Lorentz Transformation for time dilation and telling me it has something useful to say about the x' coordinate of either A' or B'. Here's the primary problem I have with it: it doesn't address position, it only addresses time. The result, when it is applied to position, strikes me as downright useless.

 

[jcsd]

Zoobyshoe:
I'll explain the exact method I used: I treated the Lorentz tensor and 4-vector postions as matrices and then simply used matrix mutiplication to perform the Lorentz transformation.

One worth pointing out s that we've all apporached the problem in slightly different ways, but still obtained equivalent results which is a strong indication in itself that we are correct.

It is worth learning 4-vector algebra as it makes problems like these extremely simple to solve and also helps to eliminate errors.

 

[Azriel]

This speed of light do-hickey

I think the answer is A.c . This is because the speed of light is measured as constant in all reference frames. you can't add a sublight speed to a light, or translight speed!

 

[jcsd]

I need for you and jcsd to justify giving me the special application of the Lorentz Transformation for time dilation and telling me it has something useful to say about the x' coordinate of either A' or B'. Here's the primary problem I have with it: it doesn't address position, it only addresses time. The result, when it is applied to position, strikes me as downright useless.

What we've done is perform the Lorentz transformation completely, to do this you must consider all four dimensions (though saying that we've cut out two because they don't change as the direction of motion is orthogonal to them, for confirmation of this in the method examine the Lorentz transformation in full), the metod I used as outlined in my last post you don't even need to consider each directrion independently.

t doesn't equal gamma, gamma is just a unitless quantity that is useful in special relativity and it comes into the equation for length contraction too.

 

[zoobyshoe]

Zoobyshoe:
I'll explain the exact method I used: I treated the Lorentz tensor and 4-vector postions as matrices and then simply used matrix mutiplication to perform the Lorentz transformation.

Fascinating!

One worth pointing out s that we've all apporached the problem in slightly different ways, but still obtained equivalent results which is a strong indication in itself that we are correct.

Somehow, I missed any statement of the results. What actual times do you get for the flashes from the train frame? Please show how you moved from - or + 346,410.15 to a value in seconds from t=0 for each flash. As I mourned to Doc Al, this number, 346,410.15 carries no meaning for me, + or -.

 

[plover]

Somehow, I missed any statement of the results. What actual times do you get for the flashes from the train frame? Please show how you moved from - or + 346,410.15 to a value in seconds from t=0 for each flash. As I mourned to Doc Al, this number, 346,410.15 carries no meaning for me, + or -.

As I said (see previous post for derivation):

So at rest the length of the train is: 2 * 1.15 * L. Now why should this be so?

The train as measured in the embankment frame has length 2L. However, the train is in motion in the embankment frame, and thus appears contracted in that frame. So, since the value we started out with measures the contracted length of the (moving) train in the embankment frame, it makes sense that in the train frame, where the train is at rest, the length - no longer being contracted - is measured to be greater.​

So if we're setting L = 300,000km then the rest length of half the train is: x' = (1.15)(300,000km) = 345,000km.

That's one significance of the number.

So far no one's calculated a value for the time of the flashes in the train frame. It would be (for the flash at B):

<br /> \beta\ =\ .5c/c\ =\ .5<br />

<br /> \gamma\ =\ \frac{1}{\sqrt{1 - (0.5)^2}}\ =\ \sqrt{4/3}\ =\ 1.15<br />

<br /> t&#039;\ =\ \frac{-\gamma\beta L}{c}<br /> \ =\ \frac{-1.15 \cdot 0.5 \cdot 300,000\ \textrm{km}}{300,000\ \textrm{km}/\textrm{s}}<br /> \ =\ -0.575\ \textrm{s}<br />​

So the event "flash at B" has coordinates

(x, t) = (L km, 0 s) = (300000 km, 0 s)​

in the embankment frame, and coordinates

(x', t') = (345000 km, -0.575 s)​

in the train frame.

 

[Phobos]

Below is an excerpt (Chapter 12) from my upcoming book...

Please don't do that again. Your post is tangential to this topic. Please start a new topic if you are looking for feedback on your writing.

 

[Doc Al]

By "at the same time" do you mean at a specific time coordinate such as T=0? Or something else?

That's exactly what I mean by "at the same time". If the train frame wishes to measure the length of something that's moving with the embankment frame, then measurements of the positions of the ends of that length must be made at the same time according to the train frame. Of course, since we already figured out length contraction, we know the answer will be L/\gamma if L is the length as measured in the embankment frame.

This is what I have understood you to be saying about this since I raised my objection. However, I haven't been able to understand from you how the answer that is produced: -346, 410.16 and + 346,410.16, contains any useful information about the when and where according to the moving frame. If I observe that the A or B event took place in the embankment frame at t=0, x=+/-300,000 km. What does a value of -/+346,410.16 tell me about the moving frame?

It tells you where the event took place according to the train frame. The events are the flashing of the lights. We know where they took place according to the embankment: at x = +/- L = +/- 300,000 km. We use the LT to find out where they took place according to the train. To find out where, use the LT equation for x': note that x' depends on both x and t. It tells us that x' = +/- 346,410.16 km. That's where these flashes take place according to the train frame. That simply means that train observers measure the location of these flashes to occur at a distance of 346,410.16 km from the origin (M'). I believe you think this is somehow wrong because you are unconsciously thinking that the two frames are somehow "lined up" at T = 0. But, no, the only point they have in common is at 0,0: the point (in space-time) where M' passes M.

According to you, this number says something useful about the when and where of event A' or B' in the moving frame. If this number actually has any use, I can't see it. In fact, to me it looks like it is saying something both incorrect and useless about these events in the A' and B' frame. These numbers seem to me to say nothing at all about the when in the A' and B' frame, and to say something outright incorrect about the position A' and B' will have in the moving frame.

To find the when, use the LT for t': note that t' depends on both x and t.

From the general direction and tone of your answers, it seems clear that you understand the difficulty I am having with these numbers. I am trying to phrase my difficulty more specifically in the hope you'll see how to solve it.

I can see that you are serious about learning this stuff, so I am happy to help. But please realize that jcsd and plover already know this stuff and they are trying to help. You are the one who needs to learn it. You will get there!

Specify which of the four you wish me to try for myself. I can use the one for x' and I can use the one for t' but there is no one of the four that gives an answer for both x' and t'. You are giving me one equation and claiming the answer is telling me both the x' and t' coordinates with one number.

Here's how it goes. An event is specified by a set of coordinates in the embankment frame: x, t. That tells where and when the event occurred in the embankment frame. Now use the LT to find where and when that same event occurred according to the train frame. You will need two equations: one that gives you x' and one that gives you t'. Here they are again:

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

jcsd gave:

\gamma=\frac {1}{\sqrt{1-\beta^2}}

where:

\beta = \frac{u}{c}

These are just the definitions of \gamma and \beta.

This is exactly the same equation as:

t=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}

No it's not. This is just a version of the the time dilation formula. It should be written as:
\Delta t=\frac{\Delta t&#039;}{\sqrt{1-\frac{v^2}{c^2}}} = \gamma \Delta t&#039;
This is a specialized formula meant for a specific use: it is not a general Lorentz transformation. (For one thing, it only applies when \Delta x&#039; = 0, which is the case for a moving clock.) This formula means: When a moving clock measures a time of \Delta t&#039;, how much time will pass according to the observer's clocks \Delta t.

I need for you and jcsd to justify giving me the special application of the Lorentz Transformation for time dilation and telling me it has something useful to say about the x' coordinate of either A' or B'. Here's the primary problem I have with it: it doesn't address position, it only addresses time. The result, when it is applied to position, strikes me as downright useless.

Apply the real Lorentz transformation as given above. Don't try to take short cuts by applying other formulas that are derived from the LT, but only apply in specific situations. (Like the "time dilation" formula or the "length contraction" formula.)

 

[zoobyshoe]

What we've done is perform the Lorentz transformation completely, to do this you must consider all four dimensions (though saying that we've cut out two because they don't change as the direction of motion is orthogonal to them, for confirmation of this in the method examine the Lorentz transformation in full),

I follow this logic without problem. y'=y and z'=z so we don't need to treat these two coordinates. We are concerned only with the two that will change from one system to the other. These two are x' and t'.

Yet, at the moment we are concerned about, t=0 and t'=0' there is nothing in the train frame coordinate for time that is different from the embankment frame coordinate for time. x' is different. At this point x is the only coordinate that is different between the two frames due to relative motion at speed .5 c.

When you and Doc Al say you are considering time and position both at once, I grasp the concept, but since the time coordinate is 0 for both frames, the net result of considering both time and location for this moment t' = 0 should be that only position, the x coordinate, is left to treat in any way.

The position of A' from M' or B' from M' should turn out to be closer together by any means of calculation than it is in the M frame. Length contracts. Instead, to my disbelief and confusion, they seem to be farther apart.

Positions A and B, are indeed, located by means of length. To say they have a coordinate is meaningless unless we locate that coordinate with repect to some point =0. This you have done, by using L, length, as the distance from M--->A and M--->B. We have designated that length to be 300,000 km.

The Lorentz transformation you gave:

\gamma = \frac{1}{\sqrt{1-\beta^2}}

gives us the figure 1.1547005 when we put .5 c in as the relative velocity of the train and the embankment.

1.1547005 times the L, 300,000, = 346,410.16

Since the coordinate system requires a length in order to define a position, it seems obvious to me that this number: 346,410.16 refers to a length. You may argue it also refers to a time because you have used the tensor Lorentz transformation, but since the time component at this point is 0, we are left only with length to transform.

Therefore, I am confused by the fact the length 300,000 km has become 346,410.16, i.e. a larger figure.

t doesn't equal gamma, gamma is just a unitless quantity that is useful in special relativity and it comes into the equation for length contraction too.

Bear the concept of context in mind. In the context of my post, the statement: t = \gamma was completely accurate.

 

[Doc Al]

Just a quick comment, as I know all will be clear when you read my previous post!

Yet, at the moment we are concerned about, t=0 and t'=0' there is nothing in the train frame coordinate for time that is different from the embankment frame coordinate for time. x' is different. At this point x is the only coordinate that is different between the two frames due to relative motion at speed .5 c.

When you and Doc Al say you are considering time and position both at once, I grasp the concept, but since the time coordinate is 0 for both frames, the net result of considering both time and location for this moment t' = 0 should be that only position, the x coordinate, is left to treat in any way.

No no no! You assume that the times are the same in both frames. Not so! They only have one point where they agree: that point where/when M' passes M. That point is 0,0 in both frames. But imagine a clock at A and a clock at B. While the embankment frame will say that the flashes occurred at t = 0 according to their clocks, the train frame sees those clocks as being out of synch! That's a key result of the "Einstein train gedanken", quantitatively explained by the LT. In the train frame, those flashes do not occur at the same time.

The position of A' from M' or B' from M' should turn out to be closer together by any means of calculation than it is in the M frame. Length contracts. Instead, to my disbelief and confusion, they seem to be farther apart.

Positions A and B, are indeed, located by means of length. To say they have a coordinate is meaningless unless we locate that coordinate with repect to some point =0. This you have done, by using L, length, as the distance from M--->A and M--->B. We have designated that length to be 300,000 km.

Again, you are talking about the length between A and M, but we are talking about the distance between events that happen at different times! Not the same thing at all!

 

[jamie]

The spped of light is independant of motion and the same for all observers.
i.e no matter how fast you travel it will always catch up with you at the same rate.
for example the speed of light is 670 million mph(not S.I units i know) if you try to outrun the light at 600million mph it will still gain on you at a rate of 670million mph

I hope this helps

 

[zoobyshoe]

Gentlemen,

I do not believe, given the amount of math a person seems to need to know to understand this situation, that I'm going to be able to grasp it.

It would be nice if I understood Relativity as you do, but since I have no actual need of it in my life, this is turning out to be more trouble than it is worth simply to satisfy a curiosity.

I think that if four intelligent, educated people working patiently over several days, can't get me over my confusion, then it is probably a sure diagnosis of a defective Relativity lobe in my brain. I'm disapointed to discover I suffer from this malady, but there is nothing to be done.

So, I am retiring from this thread. I would like to thank you all for your efforts.

Zooby

 

[jcsd]

Doc Al makes a vital point t= 0 and t'=0 are not equivalent only (x,t) and (x',t') are equivalent that is because we are dealing with two different co-ordinate systems.

imagine the transformation of a co-ordiante system on a piece of graphing paper:

x' = x + 5

y' = y + 1

The Lorentz tranfomration is simlair to this, by changing refernce frames we are just choosing a new co-ordinarte system in spacetime. I choose the above example for simplicity but in actual fact the Lorentz transformation is more like rotating the axes about a point and you won't go far wrong as visualizng the Lorentz transformation as a rotation about a point in 4-dimensional spacetime.

 

[Doc Al]

book recommendations

I do not believe, given the amount of math a person seems to need to know to understand this situation, that I'm going to be able to grasp it.

To derive and understand the LT and the famous SR effects (time dilation, length contraction, and simultaneity) all you need is a bit of algebra. (Forget about 4-vectors and tensors... for now. )

I think that if four intelligent, educated people working patiently over several days, can't get me over my confusion, then it is probably a sure diagnosis of a defective Relativity lobe in my brain. I'm disapointed to discover I suffer from this malady, but there is nothing to be done.

I think part* of the problem is that you are jumping into the middle of a problem before learning the basics. What I would recommend--the next time the relativity bug bites you--is to pick up a book geared to the beginner. One that takes you step by step, anticipates student misconceptions and thinking traps, and is focused on teaching the basics, not covering everything there is to know about SR. Here are a few that I like:

Space and Time in Special Relativity by N. David Mermin. This is probably the easiest place to start.
Space Time Physics by E. Taylor and J. Wheeler. A classic.
A Traveler's Guide to Space Time by Thomas Moore. I only recently discovered this one (even though it's 10 years old), but I like it.​

Check these out in your library and see if one suits your learning style. (Every one will require work.) Based on the questions you'ved asked, I recommend Mermin.

Good luck to you. We'll be here when you're ready.


* another part of the problem is that each of us "helpers" is approaching the problem slightly differently--that can only add to the confusion. Some of us are so used to doing these kinds of problems, that it is hard to remember how a complete SR novice feels. There are "advanced" (not really, but I'm sure it looks that way) methods for calculating SR effects--using 4-vector algebra and tensors--that make these problems trivial. But you don't need that to do these problems or to learn the basics.

 

[jcsd]

Yep, I should of said that 4-vectors and tensors aren't necessary in anyway to understanding SR, they're just useful tools (infact SR is a good way of introducing tensors).

Zoobyshoe SR is not beyond you, infact it's a very logical and easy to follow theory provided you follow it correctly.

 

[Nereid]

* another part of the problem is that each of us "helpers" is approaching the problem slightly differently--that can only add to the confusion. Some of us are so used to doing these kinds of problems, that it is hard to remember how a complete SR novice feels. There are "advanced" (not really, but I'm sure it looks that way) methods for calculating SR effects--using 4-vector algebra and tensors--that make these problems trivial. But you don't need that to do these problems or to learn the basics.

Another challenge, not necessarily in this thread, is the somewhat disjointed nature of the 'conversation', as well as an unavoidable reliance on a somewhat limiting medium (PF is great for in many ways, but real-time discussion of good diagrams, for example, isn't one of them). There's also - necessarily - a limitation in being able to match the 'teaching' to each learner's own best learning styles.

 

[plover]

I was trying to work out a way to set up a list of agreed upon points and assumptions for the thread so people wouldn't have to construct a different context for each poster, but I never finished writing up the suggestions. Plus, of course, there's no real support for that kind of thing in the forum software.

I agree with jcsd, there's no reason this should be beyond you. The scenario that's been worked out here may have become too tangled to be useful at the moment, but writers have come up with many different examples to illustrate the foundations of SR. Working through the logic of one of these other examples might be all that's necessary to bring SR into focus for you.

If you are a visual learner, the Einstein essay, good as it is, is probably not the best place to start as it is so sparsely illustrated. More recent treatments tend to have a lot of diagrams to show the different steps of the logic, e.g. the Wheeler book mentioned above does this. I'm not familiar with the Mermin book, but Mermin is a terrific writer - so even without Doc Al's recommendation I would expect a treatment of these issues by him to be worth a look.

 

[Richard Harris]

Hmmm, I see I'm not needed anymore, I got to page three, and i saw what i think then was the aspect that zoobyshoe seemed to miss.

I came up with the following, before it got to all that mathematics ( i couldn't follow as clearly as i'd have liked)

A & B are the spaceships. to maintain compatability, just drop A and only look at references to B

Lightning flashes at points A & B when Rocket is at point M as seen from Observer at point X ( at rest with A & B, located Parallel to M, at some distance to A-M-B)

the time for light from A & B to reach point M is 1 second.

Rocket is moving away from point A, towards point B at .5c (stated because it's the most important fact.

in 1 second, light from B has passed the rocket and reaches point M
in 1 second, Light from A has reached point M, the rocket is another .5 Second away in terms of distance to travel at light speed, so the Observer on the rocket has seen the flash from B, but has yet to see the flash from A.

from this I conclude the observer sees the flash from B at .66 Seconds from M and Flash from A at 1.66 seconds from M

while the Observer at X see's both flashes after one second (slightly longer than one second, as X is part of a Rightangle triangle AMX)

 

[Richard Harris]

Forgot to add, an observer at point X would see the Flashes at the same instant s/he saw the rocket halfway between M & B. I got the impression Zoo was thinking they all saw the flashes when the rocket was at point M.

the lightning hit when both saw the rocket at point M, not both saw the light when the rocket was at point M.

 

[Chronos]

Ouch. I tried to follow the needlepoint in this thread. My head hurts. Simplification. When a photon is emitted, everything in its path 'freezes'. No matter what speed the receiver is traveling, the photon will reach it in exactly d/c seconds [according to the receiver's clock]. Call it time dilation, length contraction, or whatever. It will get there right on 'time'.