is $LT$ a Lorentz invariant?

We know the relations of time dilation and length contraction $$L=\frac{L_0}\gamma\\ T=\gamma T_0$$ If we multiply them together, we get $$LT=L_0T_0.$$ This holds for all $$L$$ and $$T.$$

So, is $$LT=L'T'$$ a Lorentz invariant? If so, why is there no talk about it at all? I hope someone clears up this point for me.

I am also very confused about length contraction and time dilation conceptually and computationally. I tried watching many videos and reading many articles about it. I can't seem to differentiate between lengths and position coordinates, but that's another question I guess.

• You do indeed, seem to have found a new Lorentz invariant, but I am not sure what its usefulness is. As for you other questions, try to think of very specific situations that you find confusing and post them as a new questions. We will do our best to explain.
– KDP
Commented Jul 20 at 15:11
• I can't resist comparing $LT$ and the well-known invariant $L^2-c^2T^2$ to other famous invariants, $E\cdot B$ and $E^2-c^2B^2$. Of course, $T$ isn't a vector, so this comparison is obviously inexact in terms of Lorentz transformations.
– J.G.
Commented Jul 20 at 18:15
• @J.G. If we combine both, $$L^2-c^2T^2=L'^2-c^2T'^2\\2icLT=2icL'T'\\(L+icT)^2=(L'+icT')^2\\ L+icT=L'+icT'.$$ Does this make any sense? Commented Jul 20 at 18:21
• Well, let me put it like this: if you define a complex vector $F=E+icB$, something wonderful happens to Maxwell's equations.
– J.G.
Commented Jul 20 at 18:34
• @J.G. What happens if you define it? Commented Jul 20 at 18:47

Consulting any text-book on special relativity, you will find the well-known fact that the four-dimensional volume element $$d^4x = c \, dt \, dx \, dy \, dz$$ is a Lorentz invariant as a consequence of $$\det \Lambda = 1$$ for a (proper) Lorentz transformation $$x^{\prime \mu} = \Lambda^\mu_{\, \nu}\, x^\nu$$, implying $$d^4 x^\prime = d^4x$$.

• Thank you, but I just want to be sure. Are the differentials here coordinate displacements or interval lengths or both? It seems important in deriving length contraction and time dilation to distinguish them. Commented Jul 20 at 15:16
• @user7777777 Take e.g. the simple Lorentz transformation $x^\prime = \gamma (x-v t)$, $c t^\prime = \gamma (c t- v x/c)$, $y^\prime = y$, $z^\prime= z$, then $dx$ corresponds to your $L_0$ and $dt$ to your $T_0$. Commented Jul 20 at 15:24
• Why is that? And what about $L$ and $T$? Commented Jul 20 at 16:00
• I don't think this answers the question. You'd have to show that $L_0T_0$ and $LT$ express the area of the same shape (up to Lorentz transformation), and I don't see how to do that. Commented Jul 20 at 17:04
• @benrg the opposite is true. This is the only possible acceptable answer to the original question; the OP only considered the boost direction, in which case the simplification by the OP is correct. However, there is no good reason for the boost direction to be nice, and thus only this answer is guaranteed by Lorentz transformations to be invariant in that general case. And so, there is no insight that the OP's "discovery" is actually useful outside of that specific case, because the general case is what is useful in physics. Commented Jul 20 at 17:09

What is $$L$$ and what is $$T$$?

In relativity, it's really important to start with "events". An event, $$E$$, is a single point in space-time. It doesn't move it's just a point described by $$(t=t_E, x=x_E)$$.

In another frame, it's described by different coordinates, $$(t'=t'_E, x'=x'_E)$$, which is not surprising.

Now when you talk about length contraction and time dilation, you have to consider the difference between 2 events. If you willy nilly just claim $$L$$ and $$T$$ without reference to events that define them: you will get nonsense.

So start with a ruler, and you measure its length. The important part of measuring length is that you measure the position of each end simultaneously. Call those $$E_1$$ and $$E_2$$ (I am not going to describe the coordinates):

$$E_1 = (0, 0)$$

$$E_2 = (0, L_0)$$

Now we can talk about $$L$$ and $$T$$:

$$L\equiv E_{2, x}-E_{1, x} = (L_0-0)=L_0$$

$$T\equiv E_{2, t}-E_{1, t} = (0-0)=0$$

So the ruler's length is $$L_0$$, aka: its proper length.

In a moving frame ($$S'$$--the primed frame is always the moving frame, and you should always introduce it after the "stationary" frame, with the caveat that there is no such thing as a moving or stationary frame, because: relativity), the observer, Bob-we do this alphabetically, Bob see $$L'$$ and $$T'$$, and the invariant is ($$c=1$$):

$$T^2 - L^2 = T'^2 - L'^2 = -L_0^2$$

So Bob gets:

$$L' = \sqrt{L_0^2 + T'^2} > L_0$$

WHAT! I though length was contracted, a now Bob, looking at the moving ruler says it's longer? But Bob wasn't looking at the ruler, he was looking at Alice measure each endpoint (note: Alice can't be in two places at once, but she uses her invite lattice of calibrated graduate students to report the time and position of all relevant events--you really need to make this abstraction to avoid the pitfalls of "what Alice at $$(t,0)$$ sees" and getting delayed light propagation involved, just don't do it: in a thought experiment, Alice, Bob, Charlie...are all omnipotent).

And regardless of Bob's non-zero velocity along the $$x$$-axis, he sees Alice measure the back of the ruler first, lets the ruler move for a while, and then she measure the front end of the ruler, so of course:

$$L' > L$$

Bob doesn't think she measured it correctly.

To figure out how Bob measures the ruler, we need to look at the world lines of it's endpoints.

In Alice's frame ($$S$$--I forgot to say that):

$$W_1 = (t, 0)$$ $$W_2 = (t, L_0)$$

If we transform that to Bob's frame (I'm not doing the math), you get:

$$W'_1 = (t', vt')$$ $$W'_2 = (t', vt' + L_0/\gamma)$$

so at any time $$t'$$, Bob measures the length to be $$L_0/\gamma$$.

It this point, you need to draw Minkowski diagrams and stare at them for a while, and every time you address a relativity problems:

Start by identifying relevant events in the rest frame, then transform the, to the moving frame. Always compare events, and remember distant events that are simultaneous in one frame are not in other frames.