# A Hidden Zero in Einstein's Simple Derivation

• I
• RandyReukauf
In summary, Einstein's equation (8) results in a completely degenerate equation where all variables are multiplied by zero. This discrepancy creates a discrepancy for the reader in terms of what is allowed by the laws of nature.
RandyReukauf
TL;DR Summary
Einstein's equations lead to a completely degenerate result that causes confusion and raises questions.
I was reading Einstein's Simple Derivation of the Lorentz Transformation which is Appendix I in his book Relativity: the Special & the General Theory. (Online copies can be found at Bartleby's and the Gutenberg Project websites.) I came across an interesting but confusing result by using the values from one equation in another equation.

In his equation (8), Einstein gives the Lorentz values for x′ and t′. In equation (8a), he states the condition that
$$x'^2−c^2t'^2=x^2−c^2t^2.$$ Substituting the Lorentz values for x′ and t′ into equation (8a) will eventually result in
$$(0)\, x^2 + 2\gamma^2\,(0)\, vtx - (0)\,c^2\, t^2 = 0.$$ (My math is below.) This is a completely degenerate result. To me, it is similar to claiming that ##1=2## because ##(0)1=(0)2##.

Assuming that my math is correct, the result creates a discrepancy for me. First, I don't believe that the laws of nature permit completely degenerate relationships. Second, it also seems unlikely that this is oversight which has lasted for over a century.

Can someone please clarify this discrepancy?

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My source is Appendix I: Simple Derivation of the Lorentz Transformation in Einstein's book Relativity: the Special and the General Theory. A completely degenerate result is obtained by substituting Einstein's equation (8) into his equation (8a). Einstein's equation (8) specifies the values of ##x'## and ##t'## as
$$x' = \frac{x - vt}{\sqrt{1 - \frac{v^2}{c^2}}}\,, \qquad t' = \frac{t - \frac{v}{c^2} x}{\sqrt{1 - \frac{v^2}{c^2}}}\,.$$ For brevity, let
$$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \implies \gamma^2 = \frac{1}{1 - \frac{v^2}{c^2}} \,.$$ Then his equation (8) can be expressed in a more compact form as
$$(A) \qquad x' = \gamma (x - vt), \qquad t' = \gamma (t - \frac{v}{c^2} x) \,.$$ Next, Einstein states that this equation satisfies the condition as given in his equation (8a),
$$x'^2 - c^2 t'^2 = x^2 - c^2 t^2 .$$

The degenerate result is obtained as follows.
First, place all terms in Einstein's equation (8a) on one side to form the new equation
$$(1) \qquad x'^2 - x^2 - c^2 (t'^2 - t^2) = 0 \,.$$ Second, substitute the values for ##x'## and ##t'## from (A) into the new equation,
$$(2) \qquad \big(\gamma (x - vt)\big)^2 - x^2 - c^2\big(\big(\gamma (t - \frac{v}{c^2} x) \big)^2 - t^2 \big)\big) = 0 \,.$$ Third, expand the squares,
$$(3) \qquad \gamma^2 (x^2 - 2vtx + v^2 t^2) - x^2 - c^2\big(\gamma^2 (t^2 - 2\frac{v}{c^2} tx + \frac{v^2}{c^4}x^2) - t^2 \big) = 0 \,.$$ Then rearrange the equation in terms of ##x^2##, ##tx##, and ##t^2##,
$$(4) \qquad (\gamma^2 - \,\gamma^2 \frac{v^2}{c^2} - 1 )\, x^2 + 2\gamma^2\,(1-1)\, vtx - (\gamma^2 - \gamma^2 \frac{v^2}{c^2} - 1 )\,c^2\, t^2 = 0 \,.$$ This leads to
$$(5) \qquad \big(\gamma^2 (1 - \frac{v^2}{c^2}) - 1 \big)\, x^2 + 2\gamma^2\,(1-1)\, vtx - \big(\gamma^2 (1 - \frac{v^2}{c^2}) - 1 \big)\,c^2\, t^2 = 0 \,.$$ Finally, substitute for ##\gamma^2##,
$$(6) \qquad \big(\,\frac{1}{1 - \frac{v^2}{c^2}} \,(1 - \frac{v^2}{c^2}) - 1 \big)\, x^2 + 2\gamma^2\,(1-1)\, vtx - \big(\,\frac{1}{1 - \frac{v^2}{c^2}}\, (1 - \frac{v^2}{c^2}) - 1 \big)\,c^2\, t^2 = 0 \,.$$ The result is a completely degenerate equation where all variables are multiplied by zero,
\begin{aligned}
(7) \qquad &(1 - 1)\, x^2 + 2\gamma^2\,(1-1)\, vtx - (1 - 1)\,c^2\, t^2 = 0 \\
\implies &(0)\, x^2 + 2\gamma^2\,(0)\, vtx - (0)\,c^2\, t^2 = 0
\,.
\end{aligned}

berkeman
Any equation can be turned into that “degenerate” result with a similar approach $$a=b$$$$a-b=b-b$$$$a-b=0$$$$b-b=0$$$$0=0$$So I am unsure what exactly is the problem.

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Demystifier, vanhees71, Vanadium 50 and 1 other person
$$1 + 1 = 2 \ \Rightarrow \ 1 + 1 -2 = 0 \ \Rightarrow \ 0 = 0$$

Demystifier, berkeman, Vanadium 50 and 1 other person
Adding the negative of a value to itself will always equal zero, ##a - a = 0##, but this is a trivial case that is not of much use.

In contrast, the case of a quadratic equation, ##ax^2 + bx + c = 0##, will not equal zero for all values of ##x## but only for the roots of the equation. The graph of the equation is a parabola, the solution of which can have up to two different values of ##x##. A degenerate case is when ##a=0## and the graph becomes a line, possibly crossing the ##x##-axis once. A completely degenerate case is ##(0)x^2 + (0)x + 0 = 0## that allows all values of ##x##.

This is what happens with Einstein's equations (8) and (8a). The equation at steps (7) and (8) of the original post is of the form $$a\, x^2 + b tx - a\,c^2\, t^2 = 0.$$ Instead of finding the roots of the equation, that is, the values of ##x## and ##t## that satisfy the equation, the coefficients ##a## and ##b## equal zero. It is like a mathematical tautology in that any answer works, even nonsense values (i.e., ##x=i, t = -i##). I find this to be a problem.

RandyReukauf said:
Adding the negative of a value to itself will always equal zero, a−a=0, but this is a trivial case
Yes.

RandyReukauf said:
First, place all terms in Einstein's equation (8a) on one side ... Second, substitute the values ...
There is nothing objectionable in either your math or the outcome.

The only objectionable thing is that you think it is objectionable. There is no equation in math or physics that cannot be manipulated similarly with a similar outcome. Give me any physics equations that you do like and I can easily do the same.

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RandyReukauf said:
I find this to be a problem.
Then you'll need to find a way that this is not a problem to you!

berkeman
RandyReukauf said:
TL;DR Summary: Einstein's equations lead to a completely degenerate result that causes confusion and raises questions.

Can someone please clarify this discrepancy?
You are missing the point of the derivation. We are looking for a transformation that preserves the invariant speed of light, and once we have seen the LT it is easy to verify that it does. Thus, all “derivations” of the LT, including this one of Einstein’s, may be considered heuristic arguments that the LT is a promising candidate; the proof is in the demonstration that this candidate does indeed preserve the invariant speed of light.

(Be aware that particular zero concern has a long and ugly history in the relativity-denying crowd. If you come across a source that advances it as a powerful refutation of relativity…. You’ve found a crackpot)

PeroK, Dale and berkeman
Nugatory said:
Critics of relativity
Ah yes - the "I'm smarter than Einstein and the 120 years of physicists that followed him."

(1) As pointed out, any equation can be twisted until it has an extraneous root.
(2) Even if the particular derivation is hogwash, which it isn't, it makes no difference to the physical world. One could have adopted an approach where one postulates the Lorentz transformation and derives the constancy of the speed of light. In fact, the world almost went down this path.
(3) SR is an observed fact. If you want to replace it with something else, it won't be Newtonian physics. It will be something that looks very much like SR where it has been tested.

PhDeezNutz, russ_watters and Dale
Nugatory said:
Be aware that particular zero concern has a long and ugly history in the relativity-denying crowd
Oh, really? That is such a patent non-issue that it surprises me that it got any traction at all.

I mean start with $$F=ma$$$$a=\ddot x$$and then subtract to get$$F-ma=0$$substitute and rearrange to get $$m(a-\ddot x)=0$$ and voila we get the “completely degenerate” $$m(0)=0$$So do we “deny” ##F=ma##?

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Demystifier
Dale said:
There is no equation in math or physics that cannot be manipulated similarly with a similar outcome. Give me any physics equations that you do like and I can easily do the same
So I can understand your claim, take equation (5) in the original post and substitute ##\gamma^2 =1## instead of the Lorentz factor. This makes the equation become
$$\big( - \frac{v^2}{c^2} \big)\, (x^2 - c^2\,t^2) + 2\,(1-1)\, vtx = 0\,.$$
For me to get zero on the left side, I set ##x^2 = c^2 \, t^2## or ##x = \pm c\,t##. How do you get zero on both sides?

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weirdoguy and PeroK
RandyReukauf said:
How do you get zero on both sides?
Solve ##\gamma^2=1## to get ##v=0##. Substitute into eq 5 to get ##0=0##

Oh for heaven's sake...

a = b
(a - b) = 0
c(a-b) = 0

Look! An extra zero! Einstein is wrong! Newton is wrong! Everybody is wrong!

PhDeezNutz, PeroK and russ_watters
Nugatory said:
You are missing the point of the derivation. We are looking for a transformation that preserves the invariant speed of light, and once we have seen the LT it is easy to verify that it does. Thus, all “derivations” of the LT, including this one of Einstein’s, may be considered heuristic arguments that the LT is a promising candidate; the proof is in the demonstration that this candidate does indeed preserve the invariant speed of light.
I think Einstein's equation (8a) is the equation that preserves the speed of light. My confusion is that it allows any value of ##x## and ##t##.

Consider the standard construct of two inertial reference frames S and S' which are properly oriented and overlap each other at time ##t = t' = 0## with S' having a positive velocity ##v## along the ##x##-axis. At time ##t=t'=0##, a light source in the S frame emits light in all directions. As I understand it, both reference frames should see a sphere of light centered upon their origins. After 1s in the S frame, the radius of the sphere is ##(1s)c##. The range of ##x## is ##\pm(1s)c##. If ##x, t## and the corresponding values of ##x', t'## are entered into Einstein's equation (8a), the equation works.

The problem I have is that if values of ##x, t## outside of one second and ##(1s)c## are used, the equation works. It works for ##x=10^{200}, t =-500## before the light is emitted and even nonsense values of ##x=i, t=\sqrt{-i}##. There is no constraint mathematically.

RandyReukauf said:
I think Einstein's equation (8a) is the equation that preserves the speed of light. My confusion is that it allows any value of x and t.
That isn’t exactly what 8a says. It says that the spacetime interval (from the origin) is invariant. This includes the special case of null spacetime intervals, but covers more than that.

And yes the spacetime interval is indeed invariant for all ##x## and all ##t##. Why do you think that is a problem?

RandyReukauf said:
The problem I have is that if values of x,t outside of one second and (1s)c are used, the equation works.
So you think the speed of light should only be ##c## at ##t=1##? Or the spacetime interval should only be invariant at ##x=t=1##? I am not sure why you think this is a problem

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PeterDonis
RandyReukauf said:
The problem I have is that if....
None of this matters. When the dust settles, we have ##x'=\gamma(x-vt)## and ##t'=\gamma(t-vx)## (using units in which ##c=1## by measuring time in seconds and distances in light-seconds so as not to clutter the equations up with factors of ##c^2##). We test these formulas, we find that they preserve the speed of light so they are the right answer, we're done.

If Einstein had replaced Appendix I of that book with
I found these formulas written on the wall in the men's room at the Zurich train station, checked them out, they worked, I left a nice note for the janitor who I think wrote them down for me
relativity would be on no less sound ground.

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Dale
Anyway, at this point I am not sure what the purpose of this thread is. You can take any equation and turn it into ##0=0##. That isn’t particularly useful, but the fact that you can do so doesn’t change the usefulness of the original equation.

So the only issue here is your “feeling” about the equation. We can’t really help you with that. We aren’t hypnotists or psychologists. We are physicists. All we can do is tell you that the equations are mathematically self consistent and experimentally validated. How you feel about it is up to you, but doesn’t change the science

PeterDonis
RandyReukauf said:
The degenerate result is obtained as follows.
First, place all terms in Einstein's equation (8a) on one side to form the new equation
$$(1) \qquad x'^2 - x^2 - c^2 (t'^2 - t^2) = 0 \,.$$
No, you can't do it that way. (Sheesh!) You should be trying to prove that the transformations satisfy this equation, not assume it.

Just take the expression ##x'^2 - c^2 t'^2##, and substitute the previous expressions for ##t'## and ##x'## into it. What do you get?

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RandyReukauf said:
First, place all terms in Einstein's equation (8a) on one side
This is the step that is both unnecessary and misleading. What do you get if you don't do this step?

RandyReukauf said:
I think Einstein's equation (8a) is the equation that preserves the speed of light. My confusion is that it allows any value of ##x## and ##t##.

Consider the standard construct of two inertial reference frames S and S' which are properly oriented and overlap each other at time ##t = t' = 0## with S' having a positive velocity ##v## along the ##x##-axis. At time ##t=t'=0##, a light source in the S frame emits light in all directions. As I understand it, both reference frames should see a sphere of light centered upon their origins. After 1s in the S frame, the radius of the sphere is ##(1s)c##. The range of ##x## is ##\pm(1s)c##. If ##x, t## and the corresponding values of ##x', t'## are entered into Einstein's equation (8a), the equation works.

The problem I have is that if values of ##x, t## outside of one second and ##(1s)c## are used, the equation works. It works for ##x=10^{200}, t =-500## before the light is emitted and even nonsense values of ##x=i, t=\sqrt{-i}##. There is no constraint mathematically.
You may be fundamentally misunderstanding what is being done here. Let's just look at the mathematics.

1) Imagine the usual ##x-y## plane, but using ##t## instead of ##y##.

2) Define a function on this plane: ##f(x, t) = x^2 - c^2t^2##, where ##c## is some constant. Any constant will do.

3) Define ##x' = \gamma(x - v t)## and ##t' = \gamma(t - \frac v {c^2} x)##, where ##v## is some constant, with ##-c < v < c## and ##\gamma## is defined in the usual way.

4) We can also define ##f(x', t') = x'^2 - c^2t'^2##.

5) Exercise for the student: prove that for all ##x, t##, we have ##f(x, t) = f(x', t')##

That's a relatively simple exercise in university-level algebra. The mathematics is clear and unimpeachable.

Now, the thing you may not like, is when this particular mathematics is applied to physics. But, the mathematics is simple. And, if Einstein hadn't applied it to spacetime, no one would have batted an eyelid. No one would be saying: "algrebra is all wrong! Everyone has been doing nonsensical mathematics since Euclid started it all."

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vanhees71
RandyReukauf said:
The problem I have is that if values of ##x, t## outside of one second and ##(1s)c## are used, the equation works.
The point is to find a transformation for all ##x, t## values, that preserves light speed, not to find specific ##x, t## values.

RandyReukauf said:
There is no constraint mathematically.
Sure there is. Try a different transformation, like the Galilean one, and check if the equation is still satisfied for all possible values of ##x, t##.

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PeroK said:
Now, the thing you may not like, is when this particular mathematics is applied to physics. But, the mathematics is simple. And, if Einstein hadn't applied it to spacetime, no one would have batted an eyelid. No one would be saying: "algrebra is all wrong! Everyone has been doing nonsensical mathematics since Euclid started it all."
Hm, special relativity for sure would have been discovered also without Einstein. I'm pretty sure, there'd be as much opposition against it, if it were discovered by Lorentz or Poincare. At least the latter was very close with the mathematics all the same. The only difference to Einstein was that they couldn't let "the aether" go ;-).

RandyReukauf said:
Instead of finding the roots of the equation, that is, the values of ##x## and ##t## that satisfy the equation,
If you have one equation for two variables ##x## and ##t##, then you cannot find the definite values of ##x## and ##t##. You can only find a general relation between ##x## and ##t##, meaning that the value of one variable is arbitrary, while only the other variable is determined by this relation.

For example, consider the equation
$$x^2+t^2-2xt=0 \;\;\; (1)$$
I claim that the solution of this equation is
$$x=t \;\;\; (2)$$
Here ##t## can be thought of as being arbitrary, while only ##x## is determined by the solution. A simple way to find this solution is to recognize that the equation (1) can be written as a full square
$$(x-t)^2=0$$
which implies ##(x-t)=0##, i.e. the solution is (2).

Now let's see what happens if I apply your way of reasoning. If I insert (2) into (1) I get
$$t^2+t^2-2tt=0$$
or
$$t(t+t-2t)=0$$
i.e.
$$t (0)=0 \;\;\; (3)$$
Now you would argue that there is something wrong with it due to degeneracy. But there is nothing wrong, (2) is really the solution of (1). The result (3) only confirms that I didn't get any contradiction with that solution.

Does it help?

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