Light is emitted by matter which is at a speed lower than \(c\). Also, Light travels only at a speed \(c=299792458\) m/s. Also, Acceleration \(a=\frac{\Delta v}{\Delta t}\). as Light can only travel at \(c\), light must achieve \(c\) in \(\Delta t=0\). if \(a=\frac{\Delta v}{0}\) then \(a=\infty\).Will it happen in this universe? or is there any medium in this universe? I request everyone to comment on this... Please Don't Ignore... Even if yours or my answer is wrong,comment on this...

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TopNewestYour assumption that \(\Delta t=0\) is right but you also assume that \(\Delta v \neq 0\) which is not right. Actually \(\Delta v\) is also equal to \(0\). This is because the initial and final speed are same and equal to \(c\). – Shubham Srivastava · 3 years ago

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mustoccur. so,\(\Delta v\neq0\). – Fahad Shihab · 3 years agoLog in to reply

\(c\), is the speed of light in a vaccum, light doesn't slow down (de-accelerate), it just travels slower in a different medium. End of story. – Yong See Foo · 3 years ago

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– Fahad Shihab · 3 years ago

i am talking about light in a vacuum. Not mediums.Log in to reply

Now another news... check this out! – Fahad Shihab · 3 years ago

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Photons can only travel at a speed of \(c\) in which beforehand, for instance when emitted from a atom; they simply didn't exist in the form of a photon. Photons theoretically have no mass as they do not interact with the Higgs field; hence cannot be accelerated. – Joel Jablonski · 3 years ago

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– Fahad Shihab · 3 years ago

OK. so, if it didn't exist in the form of a photon, it must have a travelling speed, that is \(c=299792458ms^{-1}\) and according to my knowledge, \(a=\frac{\Delta v}{\Delta t}\) as acceleration is a funcion of time, not mass. So, even if its mass is zero, it should have an acceleration. Can't it be?Log in to reply

The acceleration of a particle is not determined by the time, it is determined by its mass or the Higgs field. I haven't studied general relativity in depth though seeming as the photon does not interact with the Higgs field, it can not change its speed. Hence, the acceleration of the photon is \(a=\frac{Δv}{Δt}\) where change in velocity is 0, hence acceleration must always be 0 no matter what time, even at t=0. – Joel Jablonski · 3 years ago

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– Fahad Shihab · 3 years ago

So, physically, light isn't visible to us. That means it is a wave and photons are imaginary. Then what is the reason for self propagation of light?Log in to reply

Rule of Mathematics, In Shape \(\frac{a}{b}\) , with b \(\neq\) 0, so there exists this possibility !\(\Delta\)\(t\) \(\neq\) 0 – Gabriel Merces · 3 years agoLog in to reply

– Fahad Shihab · 3 years ago

in relativity, it is said that light can't travel at a speed lower than \(c\)Log in to reply

– Alex Dias · 3 years ago

It has to be a time for less than Be, After All Denominator must be \(\neq\) of 0Log in to reply

– Fahad Shihab · 3 years ago

that's why i am saying that even light can travel at speeds lower than \(c\)Log in to reply