# Question on special relativity

I am trying to learn special relativity. If we consider two inertial reference frames with spacetime co-ordinates $$(t,x,y,z)$$ and $$(t',x',y',z')$$ and let there be 2 observers who measure the speed of light in their respective frames. Observer 1 would measure the speed of light as $$\frac{|\Delta x|}{|\Delta t|}$$ and same goes for observer 2 except they measure the speed of light as $$\frac{|\Delta x'|}{|\Delta t'|}$$.

Assuming the speed of light is constant for every observer (the postulate of relativity), this gives me (remember speed = distance / time):

$$\frac{|\Delta x|}{|\Delta t|} = \frac{|\Delta x'|}{|\Delta t'|} = c$$ (assumption 1)

Further assuming that the spatial distance between two points is given by Euclidean norm (assumption 2), I get:

$$\frac{\sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}}{|\Delta t|} = \frac{\sqrt{(\Delta x')^2 + (\Delta y')^2 + (\Delta z')^2}}{|\Delta t'|} = c$$ (master equation)

with a slight abuse of notation (in above $$\Delta x$$ denotes just the delta in the x-coordinate). Above equation can be equivalently stated as:

$$c\Delta t - (\sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}) = c\Delta t' - (\sqrt{(\Delta x')^2 + (\Delta y')^2 + (\Delta z')^2}) = 0$$ (1)

($$|\Delta t| = \Delta t$$ by causality). However, the relativity literature uses the equation:

$$(c\Delta t)^2 - ((\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2) = (c\Delta t')^2 - ((\Delta x')^2 + (\Delta y')^2 + (\Delta z')^2) = 0$$ (2)

as a starting point to derive the Lorentz transformation matrix.

My question(s):

1. What would the theory of relativity look like if we take (1) as the starting point?
2. Is there anything wrong in taking (1) as the starting point?

I don't think (1) and (2) are equivalent since:

$$a^2 - b^2 = c^2 - d^2$$ (form of 2) does not mean that $$a - b = c - d$$ (form of 1) (and vice-versa)

On the other hand both (1) and (2) can be derived from the master equation.

I would say (1) is equivalent to the master equation (meaning you can take the master equation and get (1) or go in reverse direction) whereas (2) is not strictly equivalent to the master equation as its got by squaring the master equation. It seems (1) provides stricter bounds on the theory.

• I don't think that using Euclidean geometry is allowed in special relativity since this theory is built on Minkowski space, these are two fundamentally different things. Commented Jul 12 at 16:55
• It doesn't matter what you call assumption 2 (Euclidean vs. Minkowski vs Charlie) as its used by both (1) and (2) Commented Jul 12 at 16:58
• there is no point in including $y$ and $z$, since you can define $x$ to be along the direction propagation. It makes the question too hard to read.
– JEB
Commented Jul 13 at 3:42
• @Polaris5744: Using Euclidean geometry in relativity theory is just fine, if you keep your head on straight and you use it for the right thing. Every space-like slice of Minkowski space, obtained by choosing a specific time $t_0$ and setting $t=t_0$, may be regarded as an isometric copy of Euclidean 3-space with $x,y,z$ coordinates. Commented Jul 13 at 14:02
• @Polaris5744 The difference between special relativity geometry and classical geometry is solely in the metric/norm (or psuedonorm). Euclid's postulates make no reference to metrics, so Minkowski is still Euclidean. Commented Jul 13 at 20:33

With some algebra, we can see eq. 1 and 2 are equivalent,

\begin{align} &c\Delta t - \sqrt{(\Delta x)^2+(\Delta y)^2+(\Delta z)^2} = 0 \;\;\;\; (1) \\ \Rightarrow\;\;\; &c\Delta t = \sqrt{(\Delta x)^2+(\Delta y)^2+(\Delta z)^2} \\ \Rightarrow\;\;\; &(c\Delta t)^2 = (\Delta x)^2+(\Delta y)^2+(\Delta z)^2 \\ \Rightarrow\;\;\;&(c\Delta t)^2 - \big((\Delta x)^2+(\Delta y)^2+(\Delta z)^2\big) = 0\;\;\;\; (2) \\ \end{align}

and similarly for the primed coordinate system. While you are correct $$a^2-b^2=c^2-d^2$$ does not imply $$a-b=c-d$$, it is true however that $$a^2-b^2=0$$ implies $$a-b=0$$ (if $$a$$ and $$b$$ are both positive real numbers), which is what is happening here.

Since the equations are equivalent, special relativity looks the same whichever equation you pick. You're really picking a different form of the same equation. e.g. $$x-2=0$$ and $$x=2$$ are "different" equations, but contain the same information. However, eq. 2 is preferred in relativity because it inspires the definition of a invariant inner product between four-vectors in spacetime. See e.g. here

EDIT: The answer from @naturallyInconsistent makes an excellent point. The time separation between two events connected by a beam of light can always be chosen to be positive for all observers. i.e. $$\Delta t > 0$$ and $$\Delta t'$$ > 0. However spacelike separated events satisfy the inequality

$$$$(c\Delta t)^2 - \big((\Delta x)^2+(\Delta y)^2+(\Delta z)^2\big) < 0$$$$

That is, the events are far apart spatially but not so far apart temporally. Depending on the observer, we may have $$\Delta t$$ positive, negative, or zero (see relativity of simultaneity). In such cases we can't take the square-root so willy-nilly. Especially if we want to define an invariant interval between all events in spacetime, whether they be spacelike, timelike, or null separated.

• If you include the fact that the expressions equate to 0 then yes 1 and 2 are equivalent but in derivations of the Lorentz transform the equal to zero fact is not used e.g., see en.wikipedia.org/wiki/…. If I use the method there and use my equation 1 instead of 2 (as starting point), I get very different results. I get D - B = 1 and A - C = 1. Try it. Commented Jul 12 at 18:35
• @morpheus I think you're forgetting the $y$ and $z$ pieces. Following the method from wiki, but using eq.1, you need to solve $ct-\sqrt{x^2+y^2+z^2}=ct'-\sqrt{x'^2+y'^2+z'^2}$ for $A$, $B$, $C$, and $D$. See even if you use $y'=y$ and $z'=z$, you can't subtract them from both sides, and the algebra becomes much messier. Commented Jul 12 at 19:39
• If you use the eq. 1, the $y$ and $z$ pieces aren't under a square-root so they can be canceled from both sides (assuming relative motion in the $x$-direction). Commented Jul 12 at 19:47
• Sorry w.r.t. your second comment (I agree with the equation in that comment) I did not get why I cannot cancel out the y with y' and z with z' using equation 1. After all y = y' and z = z' and that's what is used to cancel them out in the wikipedia derivation. Could you perhaps post a derivation using equation 1 as part of your answer? That would convince me that the two equations give the same result. thanks. Commented Jul 12 at 20:03
• Oh I think I see it how they are not cancelling if we use (1) even if y=y' and z=z'. I had to write on a piece of paper. But... if we assume a plane wave that is travelling in the x (and x') direction, then y, y', z, z' are all 0... Commented Jul 12 at 20:09

No, there is no way to get a satisfactory theory using $$\tag1(c\Delta t)-\sqrt{(\Delta x)^2+(\Delta y)^2+(\Delta z)^2}=(c\Delta t^\prime)-\sqrt{(\Delta x^\prime)^2+(\Delta y^\prime)^2+(\Delta z^\prime)^2}$$ for the extremely simple reason that $$\Delta t$$ or $$\Delta t^\prime$$ are allowed to be negative, and then you cannot get the equality to work out.

Lorentz transforms can change the sign of $$\Delta t$$ and so you cannot just assume that $$|\Delta t|=\Delta t$$.

Note that only for light does the interval equal to zero; for other measurements, e.g. for massive particles, the interval can be positive or negative, with physical meanings for each case. So you cannot just take the square root, because then you will get complex numbers and will need to assign meaning to them too.

• Your argument is circular. The Lorentz transformation has been derived (and is very much based) on the assumption that the variables can both be positive and negative. Commented Jul 13 at 17:46

Using $$\Delta$$anything is going to increase confusion.

Relativity is about measuring the same events in different coordinates. So for light propagating a distance $$L$$, you have 2 events:

Emission (Tx):

$$T = (0, 0)$$

and detections (Rx):

$$R = (L/c, L)$$

Now you can write the speed as:

$$v = \frac{R_x-T_x}{R_t-T_t} = \frac L {L/c} = c$$

A transformation to another coordinates system boosted by $$v$$ is:

$$T' = (0, 0)'$$

(where the prime indicates the coordinates are in the 'moving' frame), and:

$$R' =\Big( \gamma(R_t-\frac{vR_x}{c^2}), \gamma(R_x-vR_t) \Big)'$$

$$R' =\Big( \gamma(L/c-\frac{vL}{c^2}), \gamma(L-vL/c) \Big)'$$ $$R' =\Big( \gamma \frac L c(1v/c), \gamma L (1-v/c) \Big)'$$

so

$$v' = \frac{\gamma L (1-v/c)}{\gamma \frac L c(1-v/c)} = c$$