Einstein's Greatest Mistake? Examining the Lorentz Transformation

[StandardsGuy]

In Einstein’s book Relativity – the special and the general Theory (authorized translation by Robert W. Lawson, University of Sheffield) in chapter XI (the Lorentz Transformation), he gives us these formulas as the transforms:

x’ = (x-vt)/sqr(1-(v^2/c^2))

y’ = y

z’ = z

t’ = t-(v/c^2) ∙x / sqr(1-(v^2/c^2))

He then says:

If in place of the law of transmission of light we had taken as our basis the tacit assumptions of the older mechanics as to the absolute character of times and lengths, then instead of the above we should have obtained the following equations:

x’ = x-vt

y’ = y

z’ = z

t’ = t

This system of equations is often termed the “Galilei transformation.”

On page 115 (the derivation of the Lorentz transforms) he states “We require to find x’ and t’ when x and t are given.” This makes it clear that we are solving for x’ in the first formula of each set.

Now let’s examine the first of the latter set of formulas with a thought experiment. We are standing beside the train track watching the train come from our left side as it moves to our right side. When the front of the train is right in front of us, we start our stop watch. After 10 seconds we signal to a friend who is running alongside the train to mark the spot where the front of the train is then. We measure it off to be 100 feet from the spot we were standing. The variable x is the position of the train in the x axis. Where we were standing was beside the zero point. We call the velocity of the train when going from left to right positive, and consistently, we call positions on the right to be positive, and left ones to be negative. The train has moved +100 feet in 10 seconds, so its velocity is +10 feet per second.

Now let’s use the formula to calculate it: x’ = x-vt = 0-(10*10) = 0-100 = -100 feet (the wrong answer).

It is obvious that the correct formula is x’ = x+vt, which would give the correct result. This is high school level physics. Is this Einstein’s greatest mistake?

The same bad formulas are shown in the appendix I (Simple derivation of the Lorentz Transformations) on page 118.

Any thoughts on this? Could this mistake have infected later math?

 

[marcusl]

One key to special relativity is understanding that simultaneity is a very subtle thing. When you say "after 10 seconds we signal to a friend", you have ignored all that Einstein showed was important--that your signal is carried at the speed of light and has a finite propagation time, that your notion of time is different than his, etc. Einstein did it correctly and you have not. (BTW, You also have some nerve to suggest that he was wrong.)

 

[StandardsGuy]

I said nothing about simultaneity. What part of the thought experiment did you not understand? If you see a mistake in my math, point it out.
BTW The following link agrees with me: http://galileo.phys.virginia.edu/classes/252/lorentztrans.html

 

[Mister T]

Now let’s use the formula to calculate it: x’ = x-vt = 0-(10*10) = 0-100 = -100 feet (the wrong answer).

Given the equation $x'=x-vt$ we can see that $x'=0$ when both $x$ and $t$ are zero. These are the coordinates of what's called the reference event. In your example it's the front of the moving train passing by the stationary observer. When $x=0$ and $t=10$ you've shown that $x'=-100$. This tells us that a point on the train, located 100 ft behind the front of the train, will be co-located with the stationary observer 10 seconds after the reference event.

 

[marcusl]

These are transforms between coordinate systems, not the position of the train in a single coordinate system. In the primed system that is traveling with the train, the point x where you are standing is at x' = -100 after 10 seconds, which is to say, the train has passed you.

 

[Ibix]

If the platform observer assigns a coordinate (x,t) to an event then the train observer assigns (x',t'). As MisterT points out, if I am stationary on the platform at x=0 then I will be increasingly far behind an observer on the train - so my x' coordinate should get more and more negative. I think you are mis-interpreting the primed coordinates.

The virginia.edu link does not agree with you. It defines the frame S' to be moving in the +x direction as viewed from S, as you did. However, it presents the transforms from S' to S, where you present the transforms from S to S'. Your presentation of the Galilean transform is $x'=x-vt$; in the virginia.edu article it is presented as $x=x'+vt'$. Given that $t=t'$, these are obviously the same, so they agree with what you characterise as Einstein's mistake.

 

[Doc Al]

Now let’s use the formula to calculate it: x’ = x-vt = 0-(10*10) = 0-100 = -100 feet (the wrong answer).

What you've calculated, by using x = 0, is the coordinate of that point on the tracks as measured by the train at t = 10. And x' does indeed equal -100, since that point is now 100 ft behind the train.

This is high school level physics. Is this Einstein’s greatest mistake?

No, it's just your misunderstanding of what you have calculated. Read about the Galilean Transformation here: Galilean Transformation

 

[Dale]

It is obvious that the correct formula is x’ = x+vt, which would give the correct result. This is high school level physics. Is this Einstein’s greatest mistake?

Insofar as a sign convention can be correct or incorrect, Einstein's convention is correct. Even if he picked your convention it would hardly constitute a mistake, just a convention.

Your convention is not the usual one.

EDIT: the convention is whether v is the velocity of the primed frame in the unprimed coordinates or the velocity of the unprimed frame in the primed coordinates. With the axes in standard configuration one is the negative of the other.

 

[StandardsGuy]

I thank you all for straightening me out on the conventions. I had only skimmed the Virginia.edu link, and didn't notice the primes were on the right (in the first example). BTW, I wasn't saying Einstein was wrong. It was a question.

 

[Stavros Kiri]

...The variable x is the position of the train in the x axis. Where we were standing was beside the zero point. ... ... The train has moved +100 feet in 10 seconds, so its velocity is +10 feet per second.

Now let’s use the formula to calculate it: x’ = x-vt = 0-(10*10) = 0-100 = -100 feet (the wrong answer).

It' s really simple (just interpret things properly):
x' = position of (front of) train in the train's frame (i.e. expected always '0')
x = position of (front of) train in our frame (goes from '0' to '100' (ft))

Thus: x' = x-vt = 100-(10*10) = 100-100 = 0, i.e. consistent!
Your mistake was that you mixed the old event (0,0) with the new one (100,10) (in our frame) [since you put x = 0 for the new position of the train instead of 100].

Remarks: 1. To avoid similar mistakes, you should try to interpret things properly (e.g. all the quantities involved).
2. In the cases of the mentioned transformations you should be looking for coordinates of point events (x,y,z,t) [or (x,t) for only one spatial axis] and how they transform between the two different coordinate systems, and not for variables or particle coordinates (unless that's what's needed).
3. There are no regular mistakes, neither in Einstein's theories nor in any other well-established textbook theories in physics (or otherwise they would have been found!). So the first thing we have to consider in similar cases is to try to find our own mistakes! (Or that's a good use of the forum.)

 

[haushofer]

Einstein's biggest mistake was to marry his own niece.

 

[Nugatory]

Einstein's biggest mistake was to marry his own niece.

(Cousin not niece?)
Although if you had asked him he might have said that it was introducing the cosmological constant...