Hey,I have question bugging me for a couple of days now.

I wanted to ask that according to Einstein's mass energy equivalence,if scientists create a warp drive allowing them to travel FTL(suppose at a speed 'c2' mentioned in formula E=mc2),which energy will they convert in while travelling at double the light speed or will they have safety measures to prevent the ship from turning into energy?Will they be stationary in the ship according to relativity?Will slowing down again turn them into matter i.e. mass?

Warp speed and allowing FTL travel are science fiction concepts. Asking for a physics explanation of how they work is a non-sequitur.

Hey,I have question bugging me for a couple of days now.

I wanted to ask that according to Einstein's mass energy equivalence,if scientists create a warp drive allowing them to travel FTL(suppose at a speed 'c2' mentioned in formula E=mc2),which energy will they convert in while travelling at double the light speed or will they have safety measures to prevent the ship from turning into energy?Will they be stationary in the ship according to relativity?Will slowing down again turn them into matter i.e. mass?

Off-topic, but IMHO important: please note that the square of the light speed is not equal to the double of the light speed. It seems to me that you might mix these.

Warp speed and allowing FTL travel are science fiction concepts. Asking for a physics explanation of how they work is a non-sequitur.

Well, even hypothetically, a warp drive moves a pocket of space, within which the ship would locally not be traveling faster than light.

 

And even if you could travel faster than light, which you can't, E=mc^2 doesn't mean that traveling faster than light would convert you into energy.

 

So it's answerable to that extent.

Hey,I have question bugging me for a couple of days now.

I wanted to ask that according to Einstein's mass energy equivalence,if scientists create a warp drive allowing them to travel FTL(suppose at a speed 'c2' mentioned in formula E=mc2),which energy will they convert in while travelling at double the light speed or will they have safety measures to prevent the ship from turning into energy?Will they be stationary in the ship according to relativity?Will slowing down again turn them into matter i.e. mass?

It is impossible for any object/particle having mass [math]m[/math] to travel at the light speed, let alone FTL. The explanation is simple:

 

[math](mc^2)^2=E^2+(pc)^2[/math]

 

where :

[math]E[/math] is the total energy

[math]p=mv[/math] is the impulse

[math]v[/math] is the speed

 

So:

 

[math](mc^2)^2=E^2+(mvc)^2[/math]

 

This means:

 

[math]v=c \sqrt{1-\frac{m^2c^4}{E^2}}<c[/math]

 

Another way to prove the above is by starting from:

 

[math]E=\frac{mc^2}{\sqrt{1-(v/c)^2}}[/math]

 

resulting into:

 

[math]v=c \sqrt{1-\frac{m^2c^4}{E^2}}<c[/math]

Edited July 23, 201312 yr by xyzt

It is impossible for any object/particle having mass [math]m[/math] to travel at the light speed, let alone FTL. The explanation is simple:

 

[math](mc^2)^2=E^2+(pc)^2[/math]

Do you not have a typo/sign error?

 

[math](mc^2)^2=E^2-(pc)^2[/math]

 

 

where :

[math]E[/math] is the total energy

[math]p=mv[/math] is the impulse

[math]v[/math] is the speed

 

So:

 

[math](mc^2)^2=E^2+(mvc)^2[/math]

 

This means:

 

[math]v=c \sqrt{1-\frac{m^2c^4}{E^2}}<c[/math]

The answer looks right but I cannot get there algebraically from either your equation or the one with the opposite sign. Could you elaborate?

 

Another way to prove the above is by starting from:

 

[math]E=\frac{mc^2}{\sqrt{1-(v/c)^2}}[/math]

 

resulting into:

 

[math]v=c \sqrt{1-\frac{m^2c^4}{E^2}}<c[/math]

My algebra can cope with this version just fine.

Do you not have a typo/sign error?

 

[math](mc^2)^2=E^2-(pc)^2[/math]

 

 

 

The answer looks right but I cannot get there algebraically from either your equation or the one with the opposite sign. Could you elaborate?

 

 

Bah, I made two mistakes, not one, here is the correct derivation:

 

[math](mc^2)^2=E^2-(pc)^2[/math]

 

where :

[math]E[/math] is the total energy

[math]p=\frac{mv}{\sqrt{1-(v/c)^2}}[/math] is the impulse

[math]v[/math] is the speed

 

So:

 

[math](mc^2)^2=E^2-(\frac{mv}{\sqrt{1-(v/c)^2}})^2[/math]

 

This means:

 

[math]v=c \sqrt{1-\frac{m^2c^4}{E^2}}<c[/math]

 

 

 

 

My algebra can cope with this version just fine.

 

Yes, the second derivation is much easier. Here is a third one:

 

[math]p=\frac{mv}{\sqrt{1-(v/c)^2}}[/math]

 

[math]v=\frac{c}{\sqrt{1+(m/p)^2}}<c[/math]

 

This derivation has the added advantage to show that [math]v=c[/math] if and only if [math]m=0[/math], i.e. only massless particles can travel at c.

Edited July 23, 201312 yr by xyzt

Many thanks.

Many thanks.

Thank you for catching the error in the first proof.