/sci/ - Science & Math

[Anonymous No.4590498]

I've been racking my brain, trying to understand relativistic light-speed, and I think I had an 'A-ha!' moment in the shower . .

Tell me if this is accurate:

Our combined speed must ALWAYS equal light-speed. If we are standing perfectly still, all of our speed is moving through the time dimension - thus, we will observe light moving away from us at 186,000 miles per second, since time is moving 'normally.'

If we suddenly started going HALF the speed of light, half of the light-speed would be dedicated towards moving through space, which would mean that time was moving half as quickly. Newtonian physics dictates that, if we move half of the speed of light, we will witness a light-beam moving away from us at 93,000 miles per second. HOWEVER, since time is moving half as fast for us, we actually witness the light moving at 93,000 miles per HALF a second . . or, 186,000 miles per second.

Is this an accurate understanding of how the speed of light remains constant?

[Anonymous No.4590508]

No.

[Anonymous No.4590522]

Have a decent introduction into special relativity:
http://www.eftaylor.com/download.html#special_relativity

[Anonymous No.4590565]

Actually don't dismiss what OP said so quickly.

Although this was retarded:
>>which would mean that time was moving half as quickly
you should have said you are moving through time half as fast.


The reason I say not to dismiss this is because when working with spacetime we take the time dimension to be "ct" which is a distance.

You can take the idea OP is saying and apply geometry and actually DERIVE the equations of special relativity. It's kinda fun.

The formal way of dealing with this however is to make it so that instead of using a length we use an invariant, which is the spacetime interval.

So that instead of s^2 = (ct)^2 + x^2 + y^2 + z^2
which is equal to (ct)^2 for all things, in the minkowski space used by special relativity we would have:
s^2 = (ct)^2 - x^2 - y^2 - z^2

This way we can distinguish between timelike, lightlike, and spacelike intervals.

[Anonymous No.4590568]

No.

[Anonymous No.4590580]

Actually don't dismiss what OP said so quickly.

Although this was retarded:
>>which would mean that time was moving half as quickly
you should have said you are moving through time half as fast.


The reason I say not to dismiss this is because when working with spacetime we take the time dimension to be "ct" which is a distance.

You can take the idea OP is saying and apply geometry and actually DERIVE the equations of special relativity. It's kinda fun.

The formal way of dealing with this however is to make it so that instead of using a length we use an invariant, which is the spacetime interval.

So that instead of s^2 = (ct)^2 + x^2 + y^2 + z^2
which is equal to (ct")^2 in all cases (where t" is the rest frame), instead we use the minkowski space used by special relativity which then gives us:
s^2 = (ct)^2 - x^2 - y^2 - z^2

This way we can distinguish between timelike, lightlike, and spacelike intervals.

[Anonymous No.4590582]

You’re onto something, OP, but no. The relation is between momentum, rest mass, and total energy.

[Anonymous No.4590597]

>>4590580
>you should have said you are moving through time half as fast.
... and that's also wrong. As you appear to know.

Yes, OP is saying things that sound like a correct explanation. But they aren't. Sort of like saying 2+2=5. He should learn relativity so he knows how to do it correctly.

[Anonymous No.4590608]

>>4590498
>If we suddenly started going HALF the speed of light, half of the light-speed would be dedicated towards moving through space
That's not true

The factor of time dilation is 1/sqrt(1-v^2/c^2), so moving at 0.5c one second for you is 1.155s from the perspective of a body at rest.

[Anonymous No.4590613]

>>4590580
continuing from this, for all of those who think the OP is a dumb idea, I'll use OP's idea for fun and show you that it works.


Take the time dimension to be (ct), which is a distance.

If everything in the universe travels at a constant speed c within this 4-dimensional spacetime then (ct") where t" is the time measured from any rest frame is the distance the object has traveled as seen from the rest frame in a time t" through this 4-dimensional space.

(ct")^2 = (ct)^2 + x^2 + y^2 + z^2
(ct")^2 = (ct)^2 + r^2
remember that r is the distance traveled through 3D space as seen by the REST frame, so the velocity as seen by the rest frame will be r/t" = v.
<span class="math">t"^{2}=t^2+\frac{r^{2}}{c^{2}}[/spoiler]
<span class="math">t^{2}=t"^{2}-\frac{r^{2}}{c^{2}}[/spoiler]
<span class="math">\frac{t^{2}}{t"^{2}}=1-\frac{v^{2}}{c^{2}}[/spoiler]
<span class="math">\frac{t}{t"}=\sqrt{1-\frac{v^{2}}{c^{2}}}[/spoiler]
<span class="math">t=\frac{t"}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/spoiler]

hmmmm... look familiar?

[Anonymous No.4590649]

>>4590597
And in case anyone got the wrong impression, it's not just the numbers that are wrong. Invariance of c requires more than just time dilation. The most important thing to understand is relative simultaneity.

>>4590613
This is more misleading than helpful. You're trying to make
<div class="math">(ct")^2=(ct)^2+x^2+y^2+z^2</div>look more like the Euclidean distance formula. But here, ct is the hypotenuse.

[Anonymous No.4590672]

>>4590649
It extends to any number of dimensions.

distance = sqrt(a^2 + b^2 + c^2 + d^2)

would be the distance in a 4-dimensional space.

[Anonymous No.4590680]

>>4590649
He applied it just fine. If you prefer you can think of it as an application of the dot product.

<span class="math">||\vec{ct"}||^{2}=\vec{ct"}\cdot\vec{ct"}[/spoiler]
<span class="math">||\vec{ct"}||=sqrt{ct^{2}+x^{2}+y^{2}+z^{2}}[/spoiler]

[Anonymous No.4590682]

>>4590649
He applied it just fine. If you prefer you can think of it as an application of the dot product.

<span class="math">||\vec{ct"}||^{2}=\vec{ct"}\cdot\vec{ct"}[/spoiler]
<span class="math">||\vec{ct"}||=\sqrt{ct^{2}+x^{2}+y^{2}+z^{2}}[/spoiler]

[Anonymous No.4590685]

>>4590672
In spacetime the formula is
<div class="math">interval = \sqrt{(ct)^2 - x^2 - y^2 - z^2}</div>

[Anonymous No.4590684]

>>4590649
He applied it just fine. If you prefer you can think of it as an application of the dot product.

<span class="math">||\vec{ct"}||^{2}=\vec{ct"}\cdot\vec{ct"}[/spoiler]
<span class="math">||\vec{ct"}||=\sqrt{(ct)^{2}+x^{2}+y^{2}+z^{2}}[/spoiler]

[Anonymous No.4590693]

>>4590685
what's your point?

I was using ct" = interval. If we substitute that into your equation it's exactly what I wrote...

[Anonymous No.4590695]

>>4590684
The dot product in spacetime doesn't look like that.

[Anonymous No.4590708]

>>4590695
Ok, I see where you are coming from. I covered that in my first post though
>>4590580

I know that is how it is formally done, with minkowski space.

But I was just saying that for fun, if we didn't use minkowski space, but just regular old 4-dimensional space without the spacetime metric g=diag(1,-1,-1,-1)
then you can use the dot product as I did and derive the results of relativity.

It's interesting no?

[Anonymous No.4590717]

>>4590693
Right, which is why I said
>This is more misleading than helpful. You're trying to make
<div class="math">(ct")^2=(ct)^2+x^2+y^2+z^2</div>>look more like the Euclidean distance formula. But here, ct is the hypotenuse.

[Anonymous No.4590728]

>>4590708
The geometries are different. There's no way to derive one from the other (*) without making an error somewhere.

(*) You can however derive Minkowski geometry from a variation on 4-D Euclid with imaginary numbers for the time coordinates.

[Anonymous No.4590735]

>>4590728
Well maybe you can't derive all the results.

But the point is, OP's idea doesn't include minkowski space, or imaginary numbers. It's just simple 4D euclidean geometry and you can use it to derive time dilation.

Just showing that what OP was saying is not a stupid idea after all. Even if it might be inconsistent with other aspects of relativity one can see that he is justified in thinking that way only knowing basic relativity when the result derived from his idea is consistent with what he knows so far.

[Anonymous No.4590744]

>>4590735
That derivation is wrong, though. In 4D Euclidean spacetime, a person who traveled around in a spaceship while his twin stayed on earth would be OLDER than the earthbound twin when they reunited.

[Anonymous No.4590749]

>>4590744
How is it wrong?

That is exactly what the final result is. t = gamma t_0

The times are different.

[Anonymous No.4590757]

>>4590744
How is it wrong?

That is exactly what the final result is. t = gamma t_0

The times are different.

[Anonymous No.4590760]

>>4590744
perhaps you didn't read that I was using t" as the rest time.

Usually it is written t_0. But that is just the equation of time dilation:
t = t_0 gamma.

[Anonymous No.4590765]

>>4590757
The correct Euclidean result involves <span class="math">\sqrt{1+v^2/c^2}[/spoiler].

It's hard to pick out the precise critical step that's wrong in your derivation because you use concepts which are incoherent without a proper explanation of what they mean. Such as the statement that everything moves through spacetime at constant speed c.

[Anonymous No.4590769]

>>4590765
I got the exact equation for time dilation.

[Anonymous No.4590778]

>>4590769
That's because you made an error. You got the equation for Minkowski spacetime. The time dilation formula for 4D Euclidean spacetime is different from the one for Minkowski spacetime.

[Anonymous No.4590792]

>>4590778
oh alright then.

Perhaps the error was that I should have used
(ct)^2 = (ct")^2 + x^2 + y^2 + z^2
instead (with the primes switched).

well it was fun while it lasted.