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I've been working through invariant spacetime interval questions recently, and I came across a question in my lecture notes where;

$$\Delta s^2=\Delta x^2 -(c\Delta t)^2 > 0 $$

Now it is clear to me that there is no frame where $\Delta x' = 0$ which I have already proven as the question required. Now, out of curiosity, I'm wondering if there is a way to determine the minimum value that $(\Delta x')^2$ can take?

I am assuming that the spacetime interval is the same in every frame, so

$$\Delta s'^2=\Delta x'^2 -(c\Delta t')^2 > 0$$

which would give

$$\Delta x'^2 > (c\Delta t')^2$$

But since $t'$ can be equal to 0, I'm not sure where to go from here. Is there anybody that can either show me how, or point me in the right direction? Any help is much appreciated.

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  • $\begingroup$ Hi J Clarke! Welcome to PSE. I suggest you to write in your profile something about you and your present interests/knowledge. This will help others to better answer your questions in the future. I will post another answer to this question shortly $\endgroup$ – magma Nov 26 '18 at 10:42
  • $\begingroup$ Thank you J Clarke for accepting my answer. You have updated your profile so I now see that you are a Mathematics student. I suggest that - if you like an answer (eg. mine and Elio's) - you also upvote it. The same goes for questions: if you think a question was interesting or at least well written it deserves an upvote. It costs nothing yet it motivates people to ask and answer as best as possible. As you can see I upvoted your question :-) $\endgroup$ – magma Nov 28 '18 at 2:14
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Let me restate the problem the way I understand it:

we have 2 events A and B separated by a space-like interval

$$\Delta s^2=\Delta x^2 -(c\Delta t)^2 > 0 $$

now, different observers will measure these 2 events A and B and come up with different $\Delta x$ and $\Delta t$, but what will be the minimum possible $\Delta x$ (if it exists) that one of these observers might measure?

Quick answer:

from $$ 0\geq-(c\Delta t)^2$$ we obtain $$ \Delta x^2\geq\Delta x^2-(c\Delta t)^2)=\Delta s^2$$

So $\Delta x^2\geq\Delta s^2$ always in a spacelike interval and therefore the minimum value that it can attain is $\Delta s^2$, for an observer that sees A and B happen simultaneously i.e. with $\Delta t=0$

Now let's try a different , more physically insightful, apprach. First a little trick which will help better visualize the situation: let's agree that all observers reset their clocks, meters,etc such that event A has coordinates (0,0) for every observer. This does not change the motion of an observer and in general the physics of any problem. So A=(0,0) for everybody, while B=(t,x) has different coordinates for different observers, but for everybody $\Delta s^2=x^2-t^2$ is the same, say $\Delta s^2=9$ ($c=1$ from now on). Every observer will draw a space-time diagram with event A at the origin (not shown) and event B appearing somewhere. If we overlay all the diagrams we get the following

Mathematica graphics

where each observer has drawn B as a different colored dot at different positions. All these dots belong to the locus $\Delta s^2=x^2-t^2=9$ so it is clear that the green observer will measure the smallest $\Delta x^2$ and $\Delta t^2=0$

PS: the light cone in the diagram is a bonus, I could not resist putting it in ;-)

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But since $t'$ can be equal to 0, I'm not sure where to go from here.

$\let\D=\Delta$ You meant $\D t'$, didn't you? You were on the right track, but went lost. Your first equation says $$\D s^2 = \D x^2 - (c\,\D t)^2 > 0\tag1$$ Then you wrote

I am assuming that the spacetime interval is the same in every frame

which is right, but wrote

$$\D s'^2 = \D x'^2 - (c\,\D t')^2 > 0$$

forgetting what you had just written: $\D s'=\D s$. So this equation should have been $$\D s^2 = \D x'^2 - (c\,\D t')^2 > 0$$ If $\D t' = 0$ we have $$\D s^2 = \D x'^2.$$ Comparing with (1) $$\D x'^2 = \D x^2 - (c\,\D t)^2$$ and this is the least $\D x'$, given $\D x$ and $\D t$. It obtains in a frame where $\D t'=0$.

There is just one point to be understood: are we sure a frame where $\D t'=0$ actually exists? The answer is yes, but can you give a proof?

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  • $\begingroup$ Luckily I had previously proven that such a frame existed, by proving that $\beta$ in $S'$ was physically meaningful using the Lorentz transformation. I did mean $\Delta t'$ in that first quote, yes! $\endgroup$ – J. Clarke Nov 27 '18 at 11:47

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