Jump to content

Recommended Posts

Posted

Acording to the special theory of relativity you would, if you wanted to fly at the speed of light, to have to have an infinate mass, or if you were massless to start with there wouldnt be a problem. Can any body give me the formula (if complicated an explanation would be nice) that shows any body with a posative rest mass would have an infinite mass if moving at the speed of light?

Posted

ive tried working out but i come to a problem

1/sqrt(1- v²/c²)

1/sqrt(1-c²/c²)

1/sqrt(1-1)

1/Sqrt 0

1/0

Infinity

M=E/gc²

M=E/inft*c2

M=0

Posted

maybe that means you have to ahve no mass in order to travel at the speed or light rather than infinite mass :confused:

Posted
ive tried working out but i come to a problem

1/sqrt(1- v²/c²)

1/sqrt(1-c²/c²)

1/sqrt(1-1)

1/Sqrt 0

1/0

Infinity

M=E/gc²

M=E/inft*c2

M=0

 

Argh!

 

1/0 doesn't equal infinity! :)

Posted

As I remember, its not a problem travelling with the speed of light if you´re mass-less (photons).

The problem lies in accelerating a body with a mass to c. As you approach the limit c, your mass goes to infinity, meaning that your momentum and energy goes to infinity as well (which of course is impossible).

 

And yes... this is sadly one of the occasions that the friendly "1/0 = inf" doesn´t work :)

Posted

sry ed84c, I didn´t read your post carefully enough so the formula I gave you was not a complete answer to your question.

You have to add:

M = g*m, where m is the restmass and M is the "reativistic mass" (which you plug in the famous E=Mc²).

so since g->inf for v->c it follows that M->inf for v->c if m>0.

 

I was so used to "Mass=Energy" that I forgot to explicitely mention it. What you worked out (mathematical discussion about correct treatment of the limit v->c let aside) is that a particle moving at lightspeed must have a restmass of 0 which is perfectly true (photons, as stated above).

Posted

yes i have worked it out trying to get E rather than m at its OK, providing of course 1/0= infinity which i am under the impression that it is, no matter what calucator it is, or what mathematian i ask.

 

This may sound like a dumb question, but, does energy not have mass in the sense if you had a theoretical bag of energy it would way something, more that energy and mass are interchangeable, because otherwise things with energy would weigh soemthing (e.g photons) which prevents them from going at the speed of light.

 

This of course shows the problem i was confused with (i was farmiliar with energy= mass), e.g. does a moving ball weigh more than a stationary one?, drawing back to the photon problem. I would quite hapily ignore this but for the fact that if I was to prove to my friends nothing with mass could move at the speed of light (and therefore photons have no mass), they would surely ask 'but if a photon has energy....'

Posted

wrote a long text and while doing it my login expired so it´s gone now ...

 

short answers:

- relativistic mass M does not have any physical meaning (dependancy of coordinate system).

- when talking about the mass of a particle one means the restmass because it´s independent of a coordinate system.

- photons have a zero restmass.

- you don´t have to divide by zero. If you take the limit(v->c) of E you´ll see that energy approaches infinity as v approaches c (differential calculus is the english name for the field of maths that deals with limits of real values, I think. Dave will certainly know that for sure).

- the given equation implies E=0 for v->c if m=0. But the equation is not defined at v=c, so it does not apply to particles moving at c.

Posted

im sure that must have been very frustrating. Well thanks for all the help you have given me, but im still a little sure on not dividing by 0 surely at the speed of light-

1/Sqrt 1-C2/C2 is dividing by 0, and its the only way to get infinity?

Posted

also can u suggest any other formulas that i may be interested in for quantum mechanics/ relativity (not too complicated).

Posted

Relativistic doppler effect: f = f0 * (c-v)/(c+v) (can't remember whether that's with an object moving away or moving towards). f = doppler shifted freq, f0 = original frequency, c and v are obvious.

Posted
[...]' date=' but im still a little sure on not dividing by 0 surely at the speed of light-

1/Sqrt 1-C2/C2 is dividing by 0, and its the only way to get infinity?[/quote']

 

The point is that no matter how much (kinetic) energy you give to your mass>0 (infinity isn´t a number) you will never reach c so you never have to divide by zero :D.

The statement "at lightspeed you have infinite mass (=energy) if your restmass m is greater zero" strictly speaking is wrong because neither is there infinite energy nor can you reach lightspeed. It just sounds cooler than "your speed will allways be <c no matter how much kinetic energy you have" because it uses two superlatives :cool:.

 

 

>> also can u suggest any other formulas that i may be interested in for

>> quantum mechanics/ relativity (not too complicated).

 

I don´t know what you are interested in and I don´t know how good your maths is.

If you´re interested in those topics try to find online-ressources about them. Don´t bother with popular scientific stuff like Hawkin´s book. That´ll only fool you into thinking you had an idea about the topic and make you discuss the topic with other hobby scientists that also lack the fundamentals. Maybe you can try going for scripts of lectures held at universities. If you don´t understand the maths involved: Look it up. You´ll have to learn it anyways.

 

Can´t provide you with any webpages because all I´ve got are in german. Maybe I´ll google around a bit.

Posted
The point is that no matter how much (kinetic) energy you give to your mass>0 (infinity isn´t a number) you will never reach c so you never have to divide by zero :D.

The statement "at lightspeed you have infinite mass (=energy) if your restmass m is greater zero" strictly speaking is wrong because neither is there infinite energy nor can you reach lightspeed. It just sounds cooler than "your speed will allways be <c no matter how much kinetic energy you have" because it uses two superlatives :cool:.

 

Basically we say that as the velocity tends to c, we have that the mass of the object will tend to infinity (from a more mathematical standpoint).

Posted

Objects with real rest mass cannot travel at or above c, any object that travels at c MUST have a rest mass of zero.

 

The inertial mass of an object in some inertial refernce frame will tend infinity as that object's speed tends to c.

 

1/0 is undefined in the real numbers, but in the extended reals it is taken as infinity. Physicists don't use the extended reals, but they are lazy and they will often say that x is infinty at y = z if x tends to infinity as y tends to z. If you get 1/0 in an equation inn special relativity you should interpret the result as unphysical and hence referring to an impossible situation.

  • 4 weeks later...
Posted

well what exactly made scientists to think (rather made us to think) that photon have no mass.is it based on theoretical formula or experimentally proven one?

 

 

(and for relativistic doppler effect i guess a sqr rt should cum enclosin c-v/c+v,right dave!

Posted
well what exactly made scientists to think (rather made us to think) that photon have no mass.is it based on theoretical formula or experimentally proven one?

 

Considering the multitude of experiments that are based on the formulae, I'd consider it experimentally proven, albeit indirectly. You can't stop a photon and try and measure a mass. But all of the scattering processes are consistent with a massless particle.

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.