How to Add Metres & Seconds for Equal ds2: Einstein's Theory

[Pedroski55]

TL;DR Summary

Albert Einstein and Richard Conn Henry confuse me!

Reading this PDF from Professor Richard Conn Henry, I am confused by the multiplication by c on page 2.

Einstein's claim is that dx2 + dy2 + dz2 − dt2 = ds2 = dx′ 2 + dy′ 2 + dz′ 2 − dt′ 2 i.e, ds is invariant.

x, y and z are metres. t has units seconds. So I would think, "We can't subtract seconds from metres , we can't subtract dt from dx."

To sidestep that, Einstein multiplied dt by c. c has units metres per second, m/s. The seconds cancel out, leaving only m. Now we can add or subtract dx and cdt.

Professor Henry assumes dy and dz = 0, then we have: dx2 - (c dt )2 = dx′ 2 - (c dt′ )2

Are the last two variables: dx2 - (c dt )2 and dx′ 2 - (c dt′ )2 still equal to ds2??

(Sorry, I don't know how to do the mark up for mathematical expressions.)

 

[malawi_glenn]

It is very common to set c = 1 in relativity, i.e. we do not measure time in seconds anymore.

 

[Dale]

TL;DR Summary: Albert Einstein and Richard Conn Henry confuse me!

Are the last two variables: dx2 - (c dt )2 and dx′ 2 - (c dt′ )2 still equal to ds2??

Yes, for the case that $dy=dz=0$

 

[paige turner]

That wouldn't matter. The point is, the units are wrong. To convert between time and space, multiply by c:

MILES/HOUR times HOURS = MILES

Real simple. You can only understand stuff by seeing the "simple."

 

[Ibix]

So I would think, "We can't subtract seconds from metres , we can't subtract dt from dx."

It is very common in relativity to choose to work in units where $c=1$ and then suppress the $c$s to save on keeping track of them. Some sources explicitly note that they are measuring distance in seconds (also known as light seconds) or time in meters (one meter of time is approximately (1/300,000,000) seconds). Some sources just say they are setting $c=1$ by "an appropriate choice of units" or something like that. It's surprising to start doing it in introductory relativity without any comment - perhaps it was mentioned in previous lecture notes, or in class?

Are the last two variables: dx2 - (c dt )2 and dx′ 2 - (c dt′ )2 still equal to ds2??

Not in general. But if all the movement you are interested in is purely in one dimension then you can decide to call that direction $x$. Then nothing has changing $y$ or $z$ coordinates, so $dy=dz=0$. So in this case, yes it is.

(Sorry, I don't know how to do the mark up for mathematical expressions.)

See our LaTeX guide. Note that there's a known bug where it won't render in preview if you are the first person using LaTeX on a page. Refreshing the page while in preview fixes that (and Dale and I have used LaTeX in our posts, so you should be good on this page anyway).

 

[Pedroski55]

Thanks everyone!

If, (big if!) I understand it correctly, using the speed of light as a standard, 1 light second = approx. 300 000 000 metres, so it is not incongruent to subtract dt from dx as both may be measured in metres?

Professor Henry says in his pdf, because a photon is light:

Einstein predicts that dτ which is the time, by the photon's own watch, of the photon's trip—from its creation on the sun to its ceasing to exist in your retina—is: zero! By its own clock, the photon never exists!

Does this also mean, the non-existent photon also goes nowhere?? You can't go far in zero time!

 

[Ibix]

1 light second = approx. 300 000 000 metres, so it is not incongruent to subtract dt from dx as both may be measured in metres?

Correct. As long as you are measuring time in seconds and distance in light seconds, or time in years and distances in light years or whatever (nanoseconds and feet works to about 2% accuracy) then you can add distances and times without a conversion factor. Care is still needed to do so in a way that makes sense physically! You can't just walk ten feet in ten billion nanoseconds and add them together to get anything meaningful, just as adding the width of something to its height isn't usually meaningful.

Einstein predicts that dτ which is the time, by the photon's own watch, of the photon's trip—from its creation on the sun to its ceasing to exist in your retina—is: zero! By its own clock, the photon never exists!

Oh dear.

You can't define proper time along the path followed by light, so talking of a photon having a clock is (unfortunately common) nonsense. You can't build a clock that travels at $c$ even in principle. So your professor has succumbed to popsci "this sounds cool (don't think about it too hard)" nonsense there, I'm afraid.

Does this also mean, the non-existent photon also goes nowhere?? You can't go far in zero time!

Clearly the light does go somewhere. Null worldlines are counterintuitive - they mean that there are distinct events in spacetime that can be connected with paths of "length" $ds^2=0$. But this doesn't mean that "photons experience zero time" or any such nonsense - just that "proper time of a photon' is not a useful tool for considering light paths. You can still use your own time measurements.

 

[Nugatory]

Einstein predicts that dτ which is the time, by the photon's own watch, of the photon's trip—from its creation on the sun to its ceasing to exist in your retina—is: zero! By its own clock, the photon never exists!

Does this also mean, the non-existent photon also goes nowhere?? You can't go far in zero time!

No. The problem is that saying “time by the photon’s own watch” is equivalent to saying “time measured in a frame in which the photon is at rest” and there is no such frame.

This question comes up often enough that there’s a FAQ for it linked from the sticky post at the top of the subforum: https://www.physicsforums.com/threads/rest-frame-of-a-photon.511170/

 

[vanhees71]

That wouldn't matter. The point is, the units are wrong. To convert between time and space, multiply by c:

MILES/HOUR times HOURS = MILES

Real simple. You can only understand stuff by seeing the "simple."

First of all in experimental physics we use SI units. Everything else is more complicated! This means if you use the SI units lengths are measured in meters and time in seconds.

Further SI units are very inconvenient in theoretical physics, particularly in relativity and quantum (field) theory. That's why one uses what's called "natural units". In classical relativistic physics, it's very convenient to measure all velocities in terms of the speed of light in a vacuum, which according to relativity is anyway only a conversion convention between the arbitrarily chosen units of time and lenth. So one sets $c=1$, and then times and lengths are measured in the same unit, which still can be arbitrarily chosen. In high-energy physics a convenient unit of length is $1 \; \text{fm}=10^{-15} \; \text{m}$. Then times are also measured in fm, i.e., the units for measuring time and length is the same in this choice of units.

If you work in relativistic quantum theory (i.e., relativistic quantum field theory) it is further convenient to also set $\hbar=1$ since anyway $\hbar$ is another conversion constant between units. Also in the new SI it's just a defined value to introduce implicitly the unit kg of mass. Then also energy, mass, and momentum are measured in the same units, and it's the inverse of the unit of length and time. You can measure these quantities thus in terms of 1/fm, but that's also inconvenient and that's why in high-energy physics one usually measure these quantities in MeV (mega electron volts) or GeV (giga electron volts). The conversion between GeV and 1/fm is given by the relation $\hbar c \simeq 0.197 \; \text{GeV} \, \text{fm}$.

 

[vanhees71]

No. The problem is that saying “time by the photon’s own watch” is equivalent to saying “time measured in a frame in which the photon is at rest” and there is no such frame.

This question comes up often enough that there’s a FAQ for it linked from the sticky post at the top of the subforum: https://www.physicsforums.com/threads/rest-frame-of-a-photon.511170/

It's also easy to understand from a dimension-analysis point of view:

You cannot define any system of units from pure classical electromagnetic fields, because there is no dimensionful quantity in this theory, which you could use to define a natural time scale, since the electromagnetic field is massless, and you have conformal invariance in addition to the usual Poincare invariance. That's why in contradistinction to massive particles there's no natural affine time parameter for "photons" (when treated errorneously as if they were simply massless point particles).

For massive particles you can define a "natural time parameter", the "proper time", which is the time as measured in an inertial rest frame of this massive particle. Since a massless particle has no such inertial rest frame, there's no way to define an invariant time parameter in a natural way. Trajectories of massless classical particles are usually parametrized with an arbitrary affine parameter, which is determined only up to an arbitrary scaling factor, i.e., in this way there's no possibility to define a "natural time scale".

 

[Dale]

The point is, the units are wrong.

No, the units are fine. Why do you think they are wrong?

It would really help if you would use LaTeX to clearly write exactly where you think the units are wrong.

Are the last two variables: dx2 - (c dt )2 and dx′ 2 - (c dt′ )2 still equal to ds2??

In SI units we have:$$ dx^2-(c\ dt)^2 \to [\mathrm{m^2}]-[(\mathrm{m\ s^{-1}\ s})^2 ]=[ \mathrm{m^2}]-[\mathrm{m^2} ]=[\mathrm{m^2}]$$
In natural units we have $$ dx^2-(c\ dt)^2 \to [L^2]-[(L T^{-1}\ T)^2 ]=[ L^2]-[L^2 ]=[L^2] $$ or $$ dx^2- dt^2 \to [L^2]-[L^2 ]=[L^2] $$ or $$ dx^2- dt^2 \to [T^2]-[T^2 ]=[T^2] $$ All of which are also fine.

 

[PeterDonis]

It is very common to set c = 1 in relativity

Yes.

i.e. we do not measure time in seconds anymore.

Sure we can. Just measure distance in light-seconds. (Or time in years and distance in light-years.) This is actually much more common in many applications than measuring time in meters (of light-travel time) and distance in meters (or choosing some other related distance unit).

 

[malawi_glenn]

Perhaps I should have written "we do not necessarily measure time in seconds anymore"

 

[Sagittarius A-Star]

(Sorry, I don't know how to do the mark up for mathematical expressions.)

As @Ibix wrote, you find below the link "LaTeX Guide". This should be used. Please ensure to check in there the chapter "Delimiting your LaTeX code".

 

[Pedroski55]

Thank you all again for your replies.

Here is a link to another PDF from Professor Richard Conn Henry.

It seems some of you think Professor Henry is mistaken, but the maths is from Einstein and Minkowski:

$d τ^2 = dt^2 − dx^2 = dt^2 − v^2dt^2 = (1 − v^2) dt^2$

When v = 1, d τ = 0. To quote Professor Henry from the above PDF:

But $d τ^2$ is necessarily positive (and that is true also for $dt^2$ ) and so v can never be greater
than 1, because if it were, our equation would say that a positive number is equal to a negative
number: which is not true. We have discovered that if Minkowski is right (and we know from a
multitude of tests—it’s not just those muons—that Minkowski and Einstein are indeed right) then
there must be a limiting velocity: nothing can move faster than 1. One? Yes, 1 !

Professor Henry did famously write to the American Academy of Physics to tell them: the universe is in our minds! Part of that was reprinted in nature.

Thanks for all the tips and advice, I hope I got the LaTex right!

 

[Dale]

It seems some of you think Professor Henry is mistaken

What precisely do you think we think he is mistaken about.

I am really confused about what you are concerned about, both in terms of units and in terms of the paper.

 

[PeterDonis]

It seems some of you think Professor Henry is mistaken

I don't think anyone in this thread has claimed that Professor Henry is mistaken. Some of us have pointed out particular statements by you that are mistaken, but you are not Professor Henry, and the fact that you reference Professor Henry's papers does not mean that Professor Henry has endorsed everything you say.

 

[Mister T]

Professor Henry says in his pdf, because a photon is light:

Einstein predicts that dτ which is the time, by the photon's own watch, of the photon's trip—from its creation on the sun to its ceasing to exist in your retina—is: zero! By its own clock, the photon never exists!

No, Einstein does not predict that. The proper time $d \tau$ is undefined for anything traveling at light speed. That's not the same thing as saying it's zero.

 

[Ibix]

It seems some of you think Professor Henry is mistaken, but the maths is from Einstein and Minkowski:

$d τ^2 = dt^2 − dx^2 = dt^2 − v^2dt^2 = (1 − v^2) dt^2$

When v = 1, d τ = 0.

I don't disagree with that. I disagree with parts of this:

Einstein predicts that dτ which is the time, by the photon's own watch, of the photon's trip—from its creation on the sun to its ceasing to exist in your retina—is: zero! By its own clock, the photon never exists!

Does this also mean, the non-existent photon also goes nowhere?? You can't go far in zero time!

I would not call $d\tau$ "the time by the photon's own watch", although it would be a reasonable statement for any object with $v<c$, for reasons already stated. I would certainly not conclude that the photon never exists, or that it goes nowhere. The problem is that the fact that $ds^2=0$ means that notions of "the time of a photon's watch" are meaningless, and you can end up with a lot of silly conclusions trying to reason from it.

 

[Sagittarius A-Star]

Here is a link to another PDF from Professor Richard Conn Henry.

It seems some of you think Professor Henry is mistaken, but the maths is from Einstein and Minkowski:

$d τ^2 = dt^2 − dx^2 = dt^2 − v^2dt^2 = (1 − v^2) dt^2$

When v = 1, d τ = 0.

This equation is valid only for $v<1$. Richard Henry sets in his derivation of this equation on page 5 of the PDF $dx'=dy'=dz'=0$, but photons are never at rest.

The general formula is
$d s^2 = dt^2 − dx^2 = dt^2 − v^2dt^2 = (1 − v^2) dt^2$.

The following statement in the PDF is self-contradictory:

Photons.never.exist! Just ask any photon!

 

[jbriggs444]

I just can't keep from thinking of this Jingle and punning c for sea.

Photons, mermaids and undersea chickens unite!

 

[Sagittarius A-Star]

This PDF contains also wrong historical information (see at the end of page 9):

Newton had said that energy was $\frac{1}{2}mv^2$

But Newton's Principia does not define kinetic energy.

Huygens, from his percussion and compound pendulum experiments, discovered that $mv^2$ was constant
...
Finally, Leibniz seized upon $mv^2$ as ‘live force’, the cause of all effect in the universe. His new dynamics was modern in all respects except that he didn’t recognize the ‘motion of the small parts’ as—heat.

Newton was not particularly interested in engines and missed seeing the significance of $mv^2$.

Source:
https://academic.oup.com/book/8337/chapter-abstract/153981880?redirectedFrom=fulltext&login=false

Vis viva (from the Latin for "living force") is a historical term used to describe a quantity similar to kinetic energy in an early formulation of the principle of conservation of energy.
...
The recalibration of vis viva to include the coefficient of a half, namely:

was largely the result of the work of Gaspard-Gustave Coriolis and Jean-Victor Poncelet over the period 1819–1839, although the present-day definition can occasionally be found earlier (e.g., in Daniel Bernoulli's texts).

Source:
https://en.wikipedia.org/wiki/Vis_viva

 

[malawi_glenn]

I just can't keep from thinking of this Jingle and punning c for sea.

Photons, mermaids and undersea chickens unite!

I am thinking of Sheldon Coopers c-men